Daily New User Report | Sandbox

ID is: 3551 Seed is: 8804

The sine rule: finding mistakes

In maths class you are learning about the sine rule. You are working on a problem in which the triangle has sides $\text{4}$ , $\text{6}$ , and $\text{7}$ , and two angles labelled $A$ and $B$ . You write down the sine rule equation for this triangle, as shown below. But your teacher walks by and tells you that there is a mistake in the equation.

Incorrect equation:

\begin{matrix} \frac{\sin B}{7} = \frac{\sin A}{4} \end{matrix}

$\begin{array}{c} \dfrac{\sin B }{7} = \dfrac{\sin A }{4} \end{array}$

What should you do to correct the equation?

Answer:

To correct the equation you should .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule includes one side of the triangle and one angle. That side and angle must be opposite to each other in the triangle. The equation in this question includes the angles $A$ and $B$ . Are the sides in the equation both opposites of $A$ and $B$ ?

STEP: Find the sides opposite to angles

A

$A$ and

B

$B$
[−1 point ⇒ 0 / 1 points left]

The sine rule is an equation which connects two sides and two angles in any triangle:

\begin{aligned} \frac{a}{\sin A} & = \frac{b}{\sin B} \\ or \frac{\sin A}{a} & = \frac{\sin B}{b} \end{aligned}

$\begin{aligned} \dfrac{a}{\sin A} & = \dfrac{b}{\sin B} \\[0.2cm] \text{ or } \quad \dfrac{\sin A}{a} & = \dfrac{\sin B}{b} \end{aligned}$

where:

$A$ and $B$ are two angles in the triangle
$a$ is the side of the triangle opposite to angle $A$
$b$ is the side of the triangle opposite to angle $B$

For this question, that means these two sides are the important ones:

Note that $4$ is not relevant because it is not opposite to either of the angles we know. And that is why the equation in this question is wrong:

\begin{matrix} \frac{\sin B}{7} = \frac{\sin A}{\underline{4}} \end{matrix}

$\begin{array}{c} \dfrac{\sin B }{7} = \dfrac{\sin A }{{\color{red}{ \underline{4} }}} \end{array}$

All four of these equations are correct for this triangle:

\begin{aligned} \frac{6}{\sin A} & = \frac{7}{\sin B} \\ or \frac{\sin A}{6} & = \frac{\sin B}{7} \\ or \frac{7}{\sin B} & = \frac{6}{\sin A} \\ or \frac{\sin B}{7} & = \frac{\sin A}{6} \end{aligned}

$\begin{aligned} \frac{6}{\sin A} & = \frac{7}{\sin B} \\[0.2cm] \text{ or } \quad \frac{\sin A}{6} & = \frac{\sin B}{7} \\[0.2cm] \text{ or } \quad \frac{7}{\sin B} & = \frac{6}{\sin A} \\[0.2cm] \text{ or } \quad \frac{\sin B}{7} & = \frac{\sin A}{6} \end{aligned}$

TIP: The equations above show that the sine rule is quite flexible. Keep this in mind when you are solving problems with the sine rule. You can simplify your work by choosing a more convenient arrangement.

To correct the equation, change the $\text{4}$ to $\text{6}$ .

Submit your answer as:

ID is: 3551 Seed is: 9182

The sine rule: finding mistakes

In maths class you are learning about the sine rule. You are working on a problem in which the triangle has sides $\text{4}$ , $\text{5}$ , and $\text{6}$ , and two angles labelled $A$ and $B$ . You write down the sine rule equation for this triangle, as shown below. But your teacher walks by and tells you that there is a mistake in the equation.

Incorrect equation:

\begin{matrix} \frac{\sin B}{6} = \frac{\sin A}{5} \end{matrix}

$\begin{array}{c} \dfrac{\sin B }{6} = \dfrac{\sin A }{5} \end{array}$

What should you do to correct the equation?

Answer:

To correct the equation you should .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule includes one side of the triangle and one angle. That side and angle must be opposite to each other in the triangle. The equation in this question includes the angles $A$ and $B$ . Are the sides in the equation both opposites of $A$ and $B$ ?

STEP: Find the sides opposite to angles

A

$A$ and

B

$B$
[−1 point ⇒ 0 / 1 points left]

The sine rule is an equation which connects two sides and two angles in any triangle:

\begin{aligned} \frac{a}{\sin A} & = \frac{b}{\sin B} \\ or \frac{\sin A}{a} & = \frac{\sin B}{b} \end{aligned}

$\begin{aligned} \dfrac{a}{\sin A} & = \dfrac{b}{\sin B} \\[0.2cm] \text{ or } \quad \dfrac{\sin A}{a} & = \dfrac{\sin B}{b} \end{aligned}$

where:

$A$ and $B$ are two angles in the triangle
$a$ is the side of the triangle opposite to angle $A$
$b$ is the side of the triangle opposite to angle $B$

For this question, that means these two sides are the important ones:

Note that $6$ is not relevant because it is not opposite to either of the angles we know. And that is why the equation in this question is wrong:

\begin{matrix} \frac{\sin B}{\underline{6}} = \frac{\sin A}{5} \end{matrix}

$\begin{array}{c} \dfrac{\sin B }{{\color{red}{ \underline{6} }}} = \dfrac{\sin A }{5} \end{array}$

All four of these equations are correct for this triangle:

\begin{aligned} \frac{5}{\sin A} & = \frac{4}{\sin B} \\ or \frac{\sin A}{5} & = \frac{\sin B}{4} \\ or \frac{4}{\sin B} & = \frac{5}{\sin A} \\ or \frac{\sin B}{4} & = \frac{\sin A}{5} \end{aligned}

$\begin{aligned} \frac{5}{\sin A} & = \frac{4}{\sin B} \\[0.2cm] \text{ or } \quad \frac{\sin A}{5} & = \frac{\sin B}{4} \\[0.2cm] \text{ or } \quad \frac{4}{\sin B} & = \frac{5}{\sin A} \\[0.2cm] \text{ or } \quad \frac{\sin B}{4} & = \frac{\sin A}{5} \end{aligned}$

TIP: The equations above show that the sine rule is quite flexible. Keep this in mind when you are solving problems with the sine rule. You can simplify your work by choosing a more convenient arrangement.

To correct the equation, change the $\text{6}$ to $\text{4}$ .

Submit your answer as:

ID is: 3551 Seed is: 1033

The sine rule: finding mistakes

In maths class you are learning about the sine rule. You are working on a problem in which the triangle has sides $\text{6}$ , $\text{7}$ , and $\text{9}$ , and two angles labelled $A$ and $B$ . You write down the sine rule equation for this triangle, as shown below. But your teacher walks by and tells you that there is a mistake in the equation.

Incorrect equation:

\begin{matrix} \frac{\sin B}{9} = \frac{\sin A}{7} \end{matrix}

$\begin{array}{c} \dfrac{\sin B }{9} = \dfrac{\sin A }{7} \end{array}$

What should you do to correct the equation?

Answer:

To correct the equation you should .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule includes one side of the triangle and one angle. That side and angle must be opposite to each other in the triangle. The equation in this question includes the angles $A$ and $B$ . Are the sides in the equation both opposites of $A$ and $B$ ?

STEP: Find the sides opposite to angles

A

$A$ and

B

$B$
[−1 point ⇒ 0 / 1 points left]

The sine rule is an equation which connects two sides and two angles in any triangle:

\begin{aligned} \frac{a}{\sin A} & = \frac{b}{\sin B} \\ or \frac{\sin A}{a} & = \frac{\sin B}{b} \end{aligned}

$\begin{aligned} \dfrac{a}{\sin A} & = \dfrac{b}{\sin B} \\[0.2cm] \text{ or } \quad \dfrac{\sin A}{a} & = \dfrac{\sin B}{b} \end{aligned}$

where:

$A$ and $B$ are two angles in the triangle
$a$ is the side of the triangle opposite to angle $A$
$b$ is the side of the triangle opposite to angle $B$

For this question, that means these two sides are the important ones:

Note that $7$ is not relevant because it is not opposite to either of the angles we know. And that is why the equation in this question is wrong:

\begin{matrix} \frac{\sin B}{9} = \frac{\sin A}{\underline{7}} \end{matrix}

$\begin{array}{c} \dfrac{\sin B }{9} = \dfrac{\sin A }{{\color{red}{ \underline{7} }}} \end{array}$

All four of these equations are correct for this triangle:

\begin{aligned} \frac{6}{\sin A} & = \frac{9}{\sin B} \\ or \frac{\sin A}{6} & = \frac{\sin B}{9} \\ or \frac{9}{\sin B} & = \frac{6}{\sin A} \\ or \frac{\sin B}{9} & = \frac{\sin A}{6} \end{aligned}

$\begin{aligned} \frac{6}{\sin A} & = \frac{9}{\sin B} \\[0.2cm] \text{ or } \quad \frac{\sin A}{6} & = \frac{\sin B}{9} \\[0.2cm] \text{ or } \quad \frac{9}{\sin B} & = \frac{6}{\sin A} \\[0.2cm] \text{ or } \quad \frac{\sin B}{9} & = \frac{\sin A}{6} \end{aligned}$

TIP: The equations above show that the sine rule is quite flexible. Keep this in mind when you are solving problems with the sine rule. You can simplify your work by choosing a more convenient arrangement.

To correct the equation, change the $\text{7}$ to $\text{6}$ .

Submit your answer as:

ID is: 1556 Seed is: 2171

Using the sine rule

Given a triangle with an angle of $\text{79}°$ and two sides with lengths $\text{8.1}$ and $\text{12.8}$ , find the missing angle, indicated by the question mark in the figure. The figure below is drawn to scale.

TIP: You can compare the answer you get to the diagram because it is drawn to scale. The angle that you want is acute, so check your answer before you submit it to make sure it agrees with the figure.

INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.

Answer: The missing angle is °

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula and go for it!

STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 4 / 5 points left]

The sine rule works for any triangle, and it relates two angles to the two sides opposite to those angles. We can use the sine rule in either of two ways:

\begin{aligned} Angles in numerators: \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ Sides in numerators: \frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} \text{Angles in numerators:} \quad\frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\[0.4cm] \text{Sides in numerators:} \quad\frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}$

Both versions will work, but we should pick the version that makes it easier to solve the question. And in this case the first version above is the better choice because we want to find an angle and it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So it will be easier to use this arrangement of the formula:

\begin{aligned} \frac{\sin \hat{A}}{a} = \frac{\sin \hat{B}}{b} \end{aligned}

$\begin{aligned} \frac{\sin \hat{A}}{a} = \frac{\sin \hat{B}}{b} \end{aligned}$

STEP: Substitute in the known values
[−1 point ⇒ 3 / 5 points left]

When using the sine rule, we must substitute in the values carefully to make sure that they are in the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Notice that we also labelled the vertices of the triangle, in case that will be helpful. In fact, the angle we want is $\hat{ C }$ . We can use that in our equation:

\begin{aligned} \frac{\sin \hat{C}}{8.1} = \frac{\sin (79 °)}{12.8} \end{aligned}

$\begin{aligned} \frac{\sin \hat{ C }}{\text{8.1}} = \frac{\sin \left( \text{79}° \right)}{ \text{12.8}} \end{aligned}$

STEP: Solve for the missing angle
[−3 points ⇒ 0 / 5 points left]

Now it is a matter of solving for $\hat{ C }$ .

\begin{aligned} \frac{\sin \hat{C}}{8.1} & = \frac{\sin (79 °)}{12.8} \\ \frac{\sin \hat{C}}{8.1} & = \frac{(0.98162...)}{12.8} \\ \sin \hat{C} & = (0.07668...) (8.1) \\ \sin \hat{C} & = 0.62118... \end{aligned}

$\begin{aligned} \frac{\sin \hat{ C }}{\text{8.1}} & = \frac{\sin \left( \text{79}° \right)}{ \text{12.8}} \\ \frac{\sin \hat{ C }}{\text{8.1}} & = \frac{ \left( \text{0.98162...} \right) }{ \text{12.8}} \\\sin \hat{ C } & = \left( \text{0.07668...} \right)\left( \text{8.1} \right) \\\sin \hat{ C } & = \text{0.62118...} \end{aligned}$

At this point we must use the inverse sine operation. Whenever we do that, we must remember the second group of solutions and the

+ 360 ° k

$+360°k$ business. The good news is that we can ignore the

+ 360 ° k

$+360°k$ term now: we know the angle must be between zero and

180 °

$180°$ because it is in a triangle. However, we cannot ignore the second solution: we must figure out if it gives us a real answer or not. (Remember that the second solution for the sine function comes from

180^{\circ} - θ

$180^{{ \circ }} - \theta$ , where

θ

$\theta$ is whatever angle you get from your calculator.)

\begin{aligned} \hat{C} & = \sin^{- 1} (0.62118...) \\ \hat{C} & = 38.40279... ° or 141.59720... ° \end{aligned}

$\begin{aligned} \hat{ C } & = \sin^{-1}\left( \text{0.62118...} \right) \\\hat{ C } & = \text{38.40279...}° \quad \text{or} \quad \text{141.59720...}° \end{aligned}$

Now we must figure out which of these answers is correct. Or are they both possible? It turns out that the first answer, $\hat{ C } \approx \text{38.4}°$ , is the only possible answer. We know this by comparing to the diagram: the angle we want is an acute angle (less than $\text{90}\text{°}$ ). So while $\text{141.6}°$ is a valid solution to the equation we solved, it is not a valid angle for the triangle in this question. (In this case, there is another problem with $\text{141.6}°$ : if we add it to the angle given in the question, we get $\text{141.6}° + 79° = \text{220.6}°$ . And the angles in a triangle cannot be larger than $\text{180}\text{°}$ .)

NOTE: It is possible that a question like this will have two solutions, or that the obtuse angle is the correct answer. For example, if the diagram was not drawn to scale, we could not assume that the angle is acute. You should not assume that the answer will always be an acute angle!

The instructions said to round to one decimal place, so the final answer is $\hat{ C } = \text{38.4}°$ .

Submit your answer as:

ID is: 1556 Seed is: 9863

Using the sine rule

Given a triangle with a known side of $\text{6.9}$ and two known angles of $54°$ and $60°$ , find the length of the side labelled with the question mark. The figure below is drawn to scale.

TIP: You can compare the answer you get to the diagram because it is drawn to scale. The side you need is a similar length to the given side

6.9

$\text{6.9}$ so check your answer before you submit it to make sure it agrees with the figure.

INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.

Answer: The length of the missing side is units.

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula and go for it!

STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 3 / 4 points left]

The sine rule works for any triangle, and it relates two angles to the two sides opposite to those angles. We can use the sine rule in either of two ways:

\begin{aligned} Angles in numerators: \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ Sides in numerators: \frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} \text{Angles in numerators:} \quad\frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\[0.4cm] \text{Sides in numerators:} \quad\frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}$

Both versions will work, but we should pick the version that makes it easier to solve the question. And in this case the second version above is the better choice because we want to find a side and it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So it will be easier to use this arrangement of the formula:

\begin{aligned} \frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} \frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}} \end{aligned}$

STEP: Substitute in the known values
[−1 point ⇒ 2 / 4 points left]

When using the sine rule, we must substitute in the values carefully to make sure that they are in the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Notice that we also labelled the vertices of the triangle, in case that will be helpful. Each pair belongs together in one fraction. Let's use $x$ to represent the side we are looking for (the one with the ' $?$ ').

\begin{aligned} \frac{x}{\sin (54 °)} = \frac{6.9}{\sin (60 °)} \end{aligned}

$\begin{aligned} \frac{x}{\sin \left(\text{54}°\right)} = \frac{ \text{6.9} }{\sin \left(\text{60}° \right)} \end{aligned}$

STEP: Solve for the missing side
[−2 points ⇒ 0 / 4 points left]

Now it is a matter of solving for $x$ .

\begin{aligned} \frac{x}{\sin (54 °)} & = \frac{6.9}{\sin (60 °)} \\ \frac{x}{\sin (54 °)} & = \frac{6.9}{(0.86602...)} \end{aligned}

$\begin{aligned} \frac{x}{\sin \left(\text{54}°\right)} & = \frac{ \text{6.9} }{\sin \left(\text{60}° \right)} \\ \frac{x}{\sin \left(\text{54}°\right)} & = \frac{ \text{6.9} }{ \left( \text{0.86602...} \right)} \end{aligned}$

Now work on the value of the fraction on the right side and multiply both sides of the equation by the

\sin (54 °)

$\sin \left(\text{54}° \right)$ to cancel the denominator on the left.

\begin{aligned} x & = (7.96743...) \cdot \sin (54 °) \\ x & = 6.44578... \\ x & \approx 6.4 \end{aligned}

$\begin{aligned} x & = \left( \text{7.96743...} \right) \cdot \sin \left(\text{54}° \right) \\x & = \text{6.44578...} \\x & \approx \text{6.4} \end{aligned}$

Presto! We have the answer. The final step rounds the answer to one decimal place, as required in the instructions.

The unknown side of the triangle has a length of $\text{6.4}$ units.

Submit your answer as:

ID is: 1556 Seed is: 4076

Using the sine rule

Given a triangle with an angle of $\text{85}°$ and two sides with lengths $\text{7.5}$ and $\text{13.9}$ , find the missing angle, indicated by the question mark in the figure. The figure below is drawn to scale.

TIP: You can compare the answer you get to the diagram because it is drawn to scale. The angle that you want is acute, so check your answer before you submit it to make sure it agrees with the figure.

INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.

Answer: The missing angle is °

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula and go for it!

STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 4 / 5 points left]

The sine rule works for any triangle, and it relates two angles to the two sides opposite to those angles. We can use the sine rule in either of two ways:

\begin{aligned} Angles in numerators: \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ Sides in numerators: \frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} \text{Angles in numerators:} \quad\frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\[0.4cm] \text{Sides in numerators:} \quad\frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}$

Both versions will work, but we should pick the version that makes it easier to solve the question. And in this case the first version above is the better choice because we want to find an angle and it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So it will be easier to use this arrangement of the formula:

\begin{aligned} \frac{\sin \hat{A}}{a} = \frac{\sin \hat{B}}{b} \end{aligned}

$\begin{aligned} \frac{\sin \hat{A}}{a} = \frac{\sin \hat{B}}{b} \end{aligned}$

STEP: Substitute in the known values
[−1 point ⇒ 3 / 5 points left]

When using the sine rule, we must substitute in the values carefully to make sure that they are in the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Notice that we also labelled the vertices of the triangle, in case that will be helpful. In fact, the angle we want is $\hat{ C }$ . We can use that in our equation:

\begin{aligned} \frac{\sin \hat{C}}{7.5} = \frac{\sin (85 °)}{13.9} \end{aligned}

$\begin{aligned} \frac{\sin \hat{ C }}{\text{7.5}} = \frac{\sin \left( \text{85}° \right)}{ \text{13.9}} \end{aligned}$

STEP: Solve for the missing angle
[−3 points ⇒ 0 / 5 points left]

Now it is a matter of solving for $\hat{ C }$ .

\begin{aligned} \frac{\sin \hat{C}}{7.5} & = \frac{\sin (85 °)}{13.9} \\ \frac{\sin \hat{C}}{7.5} & = \frac{(0.99619...)}{13.9} \\ \sin \hat{C} & = (0.07166...) (7.5) \\ \sin \hat{C} & = 0.53751... \end{aligned}

$\begin{aligned} \frac{\sin \hat{ C }}{\text{7.5}} & = \frac{\sin \left( \text{85}° \right)}{ \text{13.9}} \\ \frac{\sin \hat{ C }}{\text{7.5}} & = \frac{ \left( \text{0.99619...} \right) }{ \text{13.9}} \\\sin \hat{ C } & = \left( \text{0.07166...} \right)\left( \text{7.5} \right) \\\sin \hat{ C } & = \text{0.53751...} \end{aligned}$

At this point we must use the inverse sine operation. Whenever we do that, we must remember the second group of solutions and the

+ 360 ° k

$+360°k$ business. The good news is that we can ignore the

+ 360 ° k

$+360°k$ term now: we know the angle must be between zero and

180 °

$180°$ because it is in a triangle. However, we cannot ignore the second solution: we must figure out if it gives us a real answer or not. (Remember that the second solution for the sine function comes from

180^{\circ} - θ

$180^{{ \circ }} - \theta$ , where

θ

$\theta$ is whatever angle you get from your calculator.)

\begin{aligned} \hat{C} & = \sin^{- 1} (0.53751...) \\ \hat{C} & = 32.51464... ° or 147.48535... ° \end{aligned}

$\begin{aligned} \hat{ C } & = \sin^{-1}\left( \text{0.53751...} \right) \\\hat{ C } & = \text{32.51464...}° \quad \text{or} \quad \text{147.48535...}° \end{aligned}$

Now we must figure out which of these answers is correct. Or are they both possible? It turns out that the first answer, $\hat{ C } \approx \text{32.5}°$ , is the only possible answer. We know this by comparing to the diagram: the angle we want is an acute angle (less than $\text{90}\text{°}$ ). So while $\text{147.5}°$ is a valid solution to the equation we solved, it is not a valid angle for the triangle in this question. (In this case, there is another problem with $\text{147.5}°$ : if we add it to the angle given in the question, we get $\text{147.5}° + 85° = \text{232.5}°$ . And the angles in a triangle cannot be larger than $\text{180}\text{°}$ .)

NOTE: It is possible that a question like this will have two solutions, or that the obtuse angle is the correct answer. For example, if the diagram was not drawn to scale, we could not assume that the angle is acute. You should not assume that the answer will always be an acute angle!

The instructions said to round to one decimal place, so the final answer is $\hat{ C } = \text{32.5}°$ .

Submit your answer as:

ID is: 3556 Seed is: 5160

Working algebraically: finding a sine ratio

Triangle $BAC$ is shown below with sides labelled $3 b$ , $20 b^{4} - 8 b^{3}$ , and $5 b - 2$ . All three of these expression are non-zero. $\angle A$ is labelled $\theta$ . The diagram is not drawn to scale.

Determine an expression for $\sin \theta$ in terms of $b$ and $\sin C$ .

Answer: $\sin \theta =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The answer should be "in terms of $b$ and $\sin C$ ". That means the answer will include both the variable $b$ and the expression $\sin C$ . You can get both of these into an equation using the sine rule.

STEP: Set up an equation using $\sin C$ and $b$
[−1 point ⇒ 2 / 3 points left]

The question says that we need an answer "in terms of $b$ and $\sin C$ ". That means we want an expression which includes both $b$ and the expression $\sin C$ . How can we do that? We can use the sine rule!

$\begin{array}{c} \dfrac{\sin \theta}{20 b^{4} - 8 b^{3}} = \dfrac{\sin C}{5 b - 2} \end{array}$

STEP: Solve for $\sin \theta$
[−1 point ⇒ 1 / 3 points left]

We want an expression for $\sin \theta$ . So we need to rearrange the equation above to make $\sin \theta$ the subject. To do this, multiply both sides of the equation by $20 b^{4} - 8 b^{3}$ .

$\begin{aligned} \dfrac{\sin \theta}{20 b^{4} - 8 b^{3}} & = \dfrac{\sin C}{5 b - 2} \\ \sin \theta & = \frac{\sin C}{5 b - 2} \cdot \left(20 b^{4} - 8 b^{3}\right) \end{aligned}$

STEP: Simplify the expression as much as possible
[−1 point ⇒ 0 / 3 points left]

The result above agrees with the question: it is an expression for $\sin \theta$ in terms of both $b$ and $\sin C$ . But we can simplify the expression on the right side: we can factorise the numerator. We can take out the common factor $4 b^{3}$ .

$\begin{aligned} \sin \theta & = \frac{\sin C}{5 b - 2} \cdot \left(20 b^{4} - 8 b^{3}\right) \\ & = \frac{\left(20 b^{4} - 8 b^{3}\right) \cdot \sin C}{5 b - 2} \\ & = \frac{\left(4 b^{3}\right)\left(5 b - 2\right) \cdot \sin C}{5 b - 2} \\ & = \left(4 b^{3}\right) \sin C \end{aligned}$

And there we have it: the expression for $\sin \theta$ in terms of $b$ and $\sin C$ is $\left(4 b^{3}\right) \sin C$ .

Submit your answer as:
Question 1 states that the three side expressions are non-zero. Why must those expressions be non-zero?

Answer:
The expressions must be non-zero .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Any time there are fractions involved in a calculation, you must worry about division by zero. Why is that?

STEP: Select the correct answer
[−1 point ⇒ 0 / 1 points left]

We need to identify the reason why the side expressions for the triangle must be non-zero. The reason is a common one: whenever there are fractions, we must be sure not to divide by zero. The solution to Question 1 uses the sine rule, and that includes fractions. So the statement "All three of these expression are non-zero" prevents division by zero. And that prevents undefined fractions.

The correct answer is that the expression must be non-zero to avoid undefined fractions.

Submit your answer as:

ID is: 3556 Seed is: 3754

Working algebraically: finding a sine ratio

Triangle $ABC$ is shown below with sides labelled $3 y - 2$ , $15 y^{5}$ , and $5 y$ . All three of these expression are non-zero. $\angle B$ is labelled $\theta$ . The diagram is not drawn to scale.

Determine an expression for $\sin \theta$ in terms of $y$ and $\sin A$ .

Answer: $\sin \theta =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The answer should be "in terms of $y$ and $\sin A$ ". That means the answer will include both the variable $y$ and the expression $\sin A$ . You can get both of these into an equation using the sine rule.

STEP: Set up an equation using $\sin A$ and $y$
[−1 point ⇒ 2 / 3 points left]

The question says that we need an answer "in terms of $y$ and $\sin A$ ". That means we want an expression which includes both $y$ and the expression $\sin A$ . How can we do that? We can use the sine rule!

$\begin{array}{c} \dfrac{\sin \theta}{15 y^{5}} = \dfrac{\sin A}{5 y} \end{array}$

STEP: Solve for $\sin \theta$
[−1 point ⇒ 1 / 3 points left]

We want an expression for $\sin \theta$ . So we need to rearrange the equation above to make $\sin \theta$ the subject. To do this, multiply both sides of the equation by $15 y^{5}$ .

$\begin{aligned} \dfrac{\sin \theta}{15 y^{5}} & = \dfrac{\sin A}{5 y} \\ \sin \theta & = \frac{\sin A}{5 y} \cdot \left(15 y^{5}\right) \end{aligned}$

STEP: Simplify the expression as much as possible
[−1 point ⇒ 0 / 3 points left]

The result above agrees with the question: it is an expression for $\sin \theta$ in terms of both $y$ and $\sin A$ . But we can simplify the expression on the right side: we can factorise the numerator. In fact, we can split the numerator such that we get a factor which matches the denominator.

$\begin{aligned} \sin \theta & = \frac{\sin A}{5 y} \cdot \left(15 y^{5}\right) \\ & = \frac{\left(15 y^{5}\right) \cdot \sin A}{5 y} \\ & = \frac{\left(3 y^{4}\right)\left(5 y\right) \cdot \sin A}{5 y} \\ & = \left(3 y^{4}\right) \sin A \end{aligned}$

And there we have it: the expression for $\sin \theta$ in terms of $y$ and $\sin A$ is $\left(3 y^{4}\right) \sin A$ .

Submit your answer as:
Question 1 states that the three side expressions are non-zero. Why must those expressions be non-zero?

Answer:
The expressions must be non-zero .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Any time there are fractions involved in a calculation, you must worry about division by zero. Why is that?

STEP: Select the correct answer
[−1 point ⇒ 0 / 1 points left]

We need to identify the reason why the side expressions for the triangle must be non-zero. The reason is a common one: whenever there are fractions, we must be sure not to divide by zero. The solution to Question 1 uses the sine rule, and that includes fractions. So the statement "All three of these expression are non-zero" prevents division by zero. And that prevents undefined fractions.

The correct answer is that the expression must be non-zero to avoid undefined fractions.

Submit your answer as:

ID is: 3556 Seed is: 7904

Working algebraically: finding a sine ratio

Triangle $CAB$ is shown below with sides labelled $3 a^{2}$ , $9 a^{4}$ , and $3 a - 4$ . All three of these expression are non-zero. $\angle A$ is labelled $\alpha$ . The diagram is not drawn to scale.

Determine an expression for $\sin \alpha$ in terms of $a$ and $\sin C$ .

Answer: $\sin \alpha =$
expression
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

The answer should be "in terms of $a$ and $\sin C$ ". That means the answer will include both the variable $a$ and the expression $\sin C$ . You can get both of these into an equation using the sine rule.

STEP: Set up an equation using $\sin C$ and $a$
[−1 point ⇒ 2 / 3 points left]

The question says that we need an answer "in terms of $a$ and $\sin C$ ". That means we want an expression which includes both $a$ and the expression $\sin C$ . How can we do that? We can use the sine rule!

$\begin{array}{c} \dfrac{\sin \alpha}{9 a^{4}} = \dfrac{\sin C}{3 a^{2}} \end{array}$

STEP: Solve for $\sin \alpha$
[−1 point ⇒ 1 / 3 points left]

We want an expression for $\sin \alpha$ . So we need to rearrange the equation above to make $\sin \alpha$ the subject. To do this, multiply both sides of the equation by $9 a^{4}$ .

$\begin{aligned} \dfrac{\sin \alpha}{9 a^{4}} & = \dfrac{\sin C}{3 a^{2}} \\ \sin \alpha & = \frac{\sin C}{3 a^{2}} \cdot \left(9 a^{4}\right) \end{aligned}$

STEP: Simplify the expression as much as possible
[−1 point ⇒ 0 / 3 points left]

The result above agrees with the question: it is an expression for $\sin \alpha$ in terms of both $a$ and $\sin C$ . But we can simplify the expression on the right side: we can factorise the numerator. In fact, we can split the numerator such that we get a factor which matches the denominator.

$\begin{aligned} \sin \alpha & = \frac{\sin C}{3 a^{2}} \cdot \left(9 a^{4}\right) \\ & = \frac{\left(9 a^{4}\right) \cdot \sin C}{3 a^{2}} \\ & = \frac{\left(3 a^{2}\right)\left(3 a^{2}\right) \cdot \sin C}{3 a^{2}} \\ & = \left(3 a^{2}\right) \sin C \end{aligned}$

And there we have it: the expression for $\sin \alpha$ in terms of $a$ and $\sin C$ is $\left(3 a^{2}\right) \sin C$ .

Submit your answer as:
Question 1 states that the three side expressions are non-zero. Why must those expressions be non-zero?

Answer:
The expressions must be non-zero .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Any time there are fractions involved in a calculation, you must worry about division by zero. Why is that?

STEP: Select the correct answer
[−1 point ⇒ 0 / 1 points left]

We need to identify the reason why the side expressions for the triangle must be non-zero. The reason is a common one: whenever there are fractions, we must be sure not to divide by zero. The solution to Question 1 uses the sine rule, and that includes fractions. So the statement "All three of these expression are non-zero" prevents division by zero. And that prevents undefined fractions.

The correct answer is that the expression must be non-zero to avoid undefined fractions.

Submit your answer as:

ID is: 3569 Seed is: 2450

Finding an angle with the sine rule

In $\Delta PQR$ , $\hat{R} = 99°$ , $RP = \text{6.9}\ \text{cm}$ , and $PQ = \text{9}\ \text{cm}$ .

Calculate the size of $\hat{Q}$ .

INSTRUCTION: Round the answer to two decimal places.

Answer:

\hat{Q} =

$\hat{Q} =$

°

$°$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation using the sine rule. The equation should look like this:

\begin{matrix} \frac{\sin \hat{Q}}{a side} = \frac{\sin (an angle)}{another side} \end{matrix}

$\begin{array}{c} \dfrac{\sin \hat{Q}}{\text{a side}} = \dfrac{\sin (\text{an angle})}{\text{another side}} \end{array}$

STEP: Write an equation using the sine rule
[−1 point ⇒ 2 / 3 points left]

We have a triangle with two sides and one angle labelled. And we are looking for an angle. This is a sine rule situation, because the sine rule always includes two sides and two angles in a triangle.

TIP:

There are two ways we can write the sine rule. Choose the way that makes it the easiest to solve the question:

Angles in numerators and sides in denominators: $\dfrac{\sin \hat{A}}{a} = \dfrac{\sin \hat{B}}{b}$
Sides in numerators and angles in denominators: $\dfrac{a}{\sin \hat{A}} = \dfrac{b}{\sin \hat{B}}$

Write the equation so that whatever you need to solve for is in the numerator.

Using the given information, we can write:

\begin{aligned} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ \frac{\sin \hat{Q}}{(6.9)} & = \frac{\sin (99 °)}{(9)} \end{aligned}

$\begin{aligned} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ \frac{\sin \hat{Q}}{(\text{6.9})} & = \frac{\sin (99°)}{(\text{9})} \end{aligned}$

STEP: Solve for

\sin \hat{Q}

$\sin \hat{Q}$
[−1 point ⇒ 1 / 3 points left]

The first step in solving for the angle is to multiply both sides of the equation by $\text{6.9}$ . This will isolate $\sin \hat{Q}$ . Then evaluate the expression on the right side.

\begin{aligned} \frac{\sin \hat{Q}}{(6.9)} & = \frac{\sin (99 °)}{(9)} \\ \sin \hat{Q} & = \frac{\sin (99 °)}{(9)} \cdot 6.9 \\ \sin \hat{Q} & = 0.75722... \end{aligned}

$\begin{aligned} \frac{\sin \hat{Q}}{(\text{6.9})} & = \frac{\sin (99°)}{(\text{9})} \\ \sin \hat{Q} & = \frac{\sin (99°)}{(\text{9})} \cdot \text{6.9} \\ \sin \hat{Q} & = \text{0.75722...} \end{aligned}$

STEP: Use the inverse sine function to find the angle
[−1 point ⇒ 0 / 3 points left]

Now we need to use the inverse sine function to isolate the angle. (Inverses undo each other, so the inverse sine function will take the sine function away from angle $\hat{Q}$ .) As always, we do this on both sides of the equation. Then use a calculator to evaluate the result.

\begin{aligned} \sin \hat{Q} & = 0.75722... \\ \sin^{- 1} (\sin \hat{Q}) & = \sin^{- 1} (0.75722...) \\ \hat{Q} & = 49.22040... \\ \hat{Q} & \approx 49.22 ° \end{aligned}

$\begin{aligned} \sin \hat{Q} & = \text{0.75722...} \\ \sin^{-1} \left(\sin \hat{Q}\right) & = \sin^{-1}\left(\text{0.75722...}\right) \\ \hat{Q} & = \text{49.22040...} \\ \hat{Q} & \approx \text{49.22}° \end{aligned}$

From our calculators we get one answer. However, the inverse sine function provides two more things that the calculator does not show us! Firstly, we get a second answer, which is $180° - \text{49.22}°$ . Secondly, the sine curve repeats every $360°$ . That means we actually get these solutions to the equation:

\begin{aligned} \hat{Q} & = 49.22 ° + 360 ° k, k \in Z \\ \hat{Q} & = 130.78 ° + 360 ° k, k \in Z \end{aligned}

$\begin{aligned} \hat{Q} & = \text{49.22}° + 360° k, \quad k \in \mathbb{Z} \\ \hat{Q} & = \text{130.78}° + 360° k, \quad k \in \mathbb{Z} \end{aligned}$

These are all the answers to the equation (there are an infinite number of them). But this question is about an angle in a triangle. This answer represents something real so the answer needs to be realistic. That means that we can ignore the $+ 360°k$ terms because the answer must be between $0°$ and $180°$ (angles cannot be negative and angles in a triangle have a sum of $180°$ ). Additionally, there is already an angle of $99°$ in the triangle, so angle $\hat{Q}$ must be less than $90°$ (because the sum of all three angles is $180°$ ). Therefore, the only answer which is acceptable is $\hat{Q} = \text{49.22}°$ .

The size of $\hat{Q}$ is $\text{49.22}°$ .

Submit your answer as:

ID is: 3569 Seed is: 6626

Finding an angle with the sine rule

In $\Delta PRQ$ , $\hat{Q} = 96°$ , $QP = \text{4.9}\ \text{cm}$ , and $PR = \text{6.6}\ \text{cm}$ .

Calculate the size of $\hat{R}$ .

INSTRUCTION: Round the answer to two decimal places.

Answer:

\hat{R} =

$\hat{R} =$

°

$°$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation using the sine rule. The equation should look like this:

\begin{matrix} \frac{\sin \hat{R}}{a side} = \frac{\sin (an angle)}{another side} \end{matrix}

$\begin{array}{c} \dfrac{\sin \hat{R}}{\text{a side}} = \dfrac{\sin (\text{an angle})}{\text{another side}} \end{array}$

STEP: Write an equation using the sine rule
[−1 point ⇒ 2 / 3 points left]

We have a triangle with two sides and one angle labelled. And we are looking for an angle. This is a sine rule situation, because the sine rule always includes two sides and two angles in a triangle.

TIP:

There are two ways we can write the sine rule. Choose the way that makes it the easiest to solve the question:

Angles in numerators and sides in denominators: $\dfrac{\sin \hat{A}}{a} = \dfrac{\sin \hat{B}}{b}$
Sides in numerators and angles in denominators: $\dfrac{a}{\sin \hat{A}} = \dfrac{b}{\sin \hat{B}}$

Write the equation so that whatever you need to solve for is in the numerator.

Using the given information, we can write:

\begin{aligned} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ \frac{\sin \hat{R}}{(4.9)} & = \frac{\sin (96 °)}{(6.6)} \end{aligned}

$\begin{aligned} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ \frac{\sin \hat{R}}{(\text{4.9})} & = \frac{\sin (96°)}{(\text{6.6})} \end{aligned}$

STEP: Solve for

\sin \hat{R}

$\sin \hat{R}$
[−1 point ⇒ 1 / 3 points left]

The first step in solving for the angle is to multiply both sides of the equation by $\text{4.9}$ . This will isolate $\sin \hat{R}$ . Then evaluate the expression on the right side.

\begin{aligned} \frac{\sin \hat{R}}{(4.9)} & = \frac{\sin (96 °)}{(6.6)} \\ \sin \hat{R} & = \frac{\sin (96 °)}{(6.6)} \cdot 4.9 \\ \sin \hat{R} & = 0.73835... \end{aligned}

$\begin{aligned} \frac{\sin \hat{R}}{(\text{4.9})} & = \frac{\sin (96°)}{(\text{6.6})} \\ \sin \hat{R} & = \frac{\sin (96°)}{(\text{6.6})} \cdot \text{4.9} \\ \sin \hat{R} & = \text{0.73835...} \end{aligned}$

STEP: Use the inverse sine function to find the angle
[−1 point ⇒ 0 / 3 points left]

Now we need to use the inverse sine function to isolate the angle. (Inverses undo each other, so the inverse sine function will take the sine function away from angle $\hat{R}$ .) As always, we do this on both sides of the equation. Then use a calculator to evaluate the result.

\begin{aligned} \sin \hat{R} & = 0.73835... \\ \sin^{- 1} (\sin \hat{R}) & = \sin^{- 1} (0.73835...) \\ \hat{R} & = 47.59165... \\ \hat{R} & \approx 47.59 ° \end{aligned}

$\begin{aligned} \sin \hat{R} & = \text{0.73835...} \\ \sin^{-1} \left(\sin \hat{R}\right) & = \sin^{-1}\left(\text{0.73835...}\right) \\ \hat{R} & = \text{47.59165...} \\ \hat{R} & \approx \text{47.59}° \end{aligned}$

From our calculators we get one answer. However, the inverse sine function provides two more things that the calculator does not show us! Firstly, we get a second answer, which is $180° - \text{47.59}°$ . Secondly, the sine curve repeats every $360°$ . That means we actually get these solutions to the equation:

\begin{aligned} \hat{R} & = 47.59 ° + 360 ° k, k \in Z \\ \hat{R} & = 132.41 ° + 360 ° k, k \in Z \end{aligned}

$\begin{aligned} \hat{R} & = \text{47.59}° + 360° k, \quad k \in \mathbb{Z} \\ \hat{R} & = \text{132.41}° + 360° k, \quad k \in \mathbb{Z} \end{aligned}$

These are all the answers to the equation (there are an infinite number of them). But this question is about an angle in a triangle. This answer represents something real so the answer needs to be realistic. That means that we can ignore the $+ 360°k$ terms because the answer must be between $0°$ and $180°$ (angles cannot be negative and angles in a triangle have a sum of $180°$ ). Additionally, there is already an angle of $96°$ in the triangle, so angle $\hat{R}$ must be less than $90°$ (because the sum of all three angles is $180°$ ). Therefore, the only answer which is acceptable is $\hat{R} = \text{47.59}°$ .

The size of $\hat{R}$ is $\text{47.59}°$ .

Submit your answer as:

ID is: 3569 Seed is: 2714

Finding an angle with the sine rule

In $\Delta QPR$ , $\hat{R} = 111°$ , $RQ = \text{3.5}\ \text{cm}$ , and $QP = \text{7}\ \text{cm}$ .

Calculate the size of $\hat{P}$ .

INSTRUCTION: Round the answer to two decimal places.

Answer:

\hat{P} =

$\hat{P} =$

°

$°$

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation using the sine rule. The equation should look like this:

\begin{matrix} \frac{\sin \hat{P}}{a side} = \frac{\sin (an angle)}{another side} \end{matrix}

$\begin{array}{c} \dfrac{\sin \hat{P}}{\text{a side}} = \dfrac{\sin (\text{an angle})}{\text{another side}} \end{array}$

STEP: Write an equation using the sine rule
[−1 point ⇒ 2 / 3 points left]

We have a triangle with two sides and one angle labelled. And we are looking for an angle. This is a sine rule situation, because the sine rule always includes two sides and two angles in a triangle.

TIP:

There are two ways we can write the sine rule. Choose the way that makes it the easiest to solve the question:

Angles in numerators and sides in denominators: $\dfrac{\sin \hat{A}}{a} = \dfrac{\sin \hat{B}}{b}$
Sides in numerators and angles in denominators: $\dfrac{a}{\sin \hat{A}} = \dfrac{b}{\sin \hat{B}}$

Write the equation so that whatever you need to solve for is in the numerator.

Using the given information, we can write:

\begin{aligned} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ \frac{\sin \hat{P}}{(3.5)} & = \frac{\sin (111 °)}{(7)} \end{aligned}

$\begin{aligned} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ \frac{\sin \hat{P}}{(\text{3.5})} & = \frac{\sin (111°)}{(\text{7})} \end{aligned}$

STEP: Solve for

\sin \hat{P}

$\sin \hat{P}$
[−1 point ⇒ 1 / 3 points left]

The first step in solving for the angle is to multiply both sides of the equation by $\text{3.5}$ . This will isolate $\sin \hat{P}$ . Then evaluate the expression on the right side.

\begin{aligned} \frac{\sin \hat{P}}{(3.5)} & = \frac{\sin (111 °)}{(7)} \\ \sin \hat{P} & = \frac{\sin (111 °)}{(7)} \cdot 3.5 \\ \sin \hat{P} & = 0.46679... \end{aligned}

$\begin{aligned} \frac{\sin \hat{P}}{(\text{3.5})} & = \frac{\sin (111°)}{(\text{7})} \\ \sin \hat{P} & = \frac{\sin (111°)}{(\text{7})} \cdot \text{3.5} \\ \sin \hat{P} & = \text{0.46679...} \end{aligned}$

STEP: Use the inverse sine function to find the angle
[−1 point ⇒ 0 / 3 points left]

Now we need to use the inverse sine function to isolate the angle. (Inverses undo each other, so the inverse sine function will take the sine function away from angle $\hat{P}$ .) As always, we do this on both sides of the equation. Then use a calculator to evaluate the result.

\begin{aligned} \sin \hat{P} & = 0.46679... \\ \sin^{- 1} (\sin \hat{P}) & = \sin^{- 1} (0.46679...) \\ \hat{P} & = 27.82614... \\ \hat{P} & \approx 27.83 ° \end{aligned}

$\begin{aligned} \sin \hat{P} & = \text{0.46679...} \\ \sin^{-1} \left(\sin \hat{P}\right) & = \sin^{-1}\left(\text{0.46679...}\right) \\ \hat{P} & = \text{27.82614...} \\ \hat{P} & \approx \text{27.83}° \end{aligned}$

From our calculators we get one answer. However, the inverse sine function provides two more things that the calculator does not show us! Firstly, we get a second answer, which is $180° - \text{27.83}°$ . Secondly, the sine curve repeats every $360°$ . That means we actually get these solutions to the equation:

\begin{aligned} \hat{P} & = 27.83 ° + 360 ° k, k \in Z \\ \hat{P} & = 152.17 ° + 360 ° k, k \in Z \end{aligned}

$\begin{aligned} \hat{P} & = \text{27.83}° + 360° k, \quad k \in \mathbb{Z} \\ \hat{P} & = \text{152.17}° + 360° k, \quad k \in \mathbb{Z} \end{aligned}$

These are all the answers to the equation (there are an infinite number of them). But this question is about an angle in a triangle. This answer represents something real so the answer needs to be realistic. That means that we can ignore the $+ 360°k$ terms because the answer must be between $0°$ and $180°$ (angles cannot be negative and angles in a triangle have a sum of $180°$ ). Additionally, there is already an angle of $111°$ in the triangle, so angle $\hat{P}$ must be less than $90°$ (because the sum of all three angles is $180°$ ). Therefore, the only answer which is acceptable is $\hat{P} = \text{27.83}°$ .

The size of $\hat{P}$ is $\text{27.83}°$ .

Submit your answer as:

ID is: 3500 Seed is: 1083

The flexibility of the sine rule

The diagram below shows a triangle, drawn to scale. One of the sides has a length of $\text{8.6}$ and another side has the length $\text{9.9}$ . The angle across from the side labelled $\text{8.6}$ is $\text{45}°$ .

Suppose you need to calculate $x$ , the angle across from the side with length $\text{9.9}$ . The two equations below are both accurate and you can use either of these versions to find $x$ . But one choice makes the solution a bit easier. Which version is the better choice?

Version 1	$\dfrac{\text{9.9}}{\sin x} = \dfrac{ \text{8.6}}{\sin \left( \text{45}° \right)}$
Version 2	$\dfrac{\sin x}{\text{9.9}} = \dfrac{\sin \left( \text{45}° \right)}{ \text{8.6}}$

Answer: The better choice is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about solving equations these two equations for $y$ :

\begin{matrix} \frac{y}{5} = \frac{7}{10} or \frac{5}{y} = \frac{10}{7} \end{matrix}

$\begin{array}{c}\frac{y}{5} = \frac{7}{10} \quad \text{or} \quad \frac{5}{y} = \frac{10}{7}\end{array}$

The equation on the left is more convenient to solve. Why?

STEP: Choose the version with the needed value in the numerator
[−1 point ⇒ 0 / 1 points left]

We can arrange the sine rule in two different ways. We can either put the angles in the numerator with the sides in the denominator, or the other way around:

\begin{aligned} _{sides in denominators:}^{Angles in numerators and} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ _{angles in denominators:}^{Sides in numerators and} \frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} ^{\text{Angles in numerators and}}_{\;\text{sides in denominators:}} \quad\frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\[0.4cm] ^{\text{Sides in numerators and}}_{\;\text{angles in denominators:}} \quad\frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}$

In both equations, $a$ and $b$ represent sides of the triangle and $\hat{A}$ and $\hat{B}$ represent angles. Solving equations with the unknown value in the numerator is usually easier than solving equations with the unknown in the denominator. So we should use the arrangement which will put the value we want to find in the numerator.

In this question, we have two versions of the sine rule for the triangle given. Both of these versions are correct. But since we want to find an angle the choice with the angles in the numerators is the better choice.

The correct choice is Version 2.

Submit your answer as:

Now solve for the missing angle, $x$ . Use the equation identified in Question 1.

INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.

Answer: The missing angle is °
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start with the equation we identified in Question 1:

$\begin{array}{c}\dfrac{\sin x}{\text{9.9}} = \dfrac{\sin \left( \text{45}° \right)}{ \text{8.6}}\end{array}$

You need to solve for $x$ .

STEP: Solve for the missing angle
[−3 points ⇒ 0 / 3 points left]

We need to find the value of $x$ , which is the missing angle. We already have the equation from Question 1. So let's solve it! Start by evaluating the right side of the equation.

$\begin{aligned} \frac{\sin x}{\text{9.9}} & = \frac{\sin \left( \text{45}° \right)}{ \text{8.6}} \\ \frac{\sin x}{\text{9.9}} & = \frac{ \left( \text{0.70710...} \right) }{ \text{8.6}} \\\frac{\sin x}{\text{9.9}} & = \text{0.08222...} \\\sin x & = \left( \text{0.08222...} \right)\left( \text{9.9} \right) \\\sin x & = \text{0.81399...} \end{aligned}$
Now we must use the inverse sine operation. This comes with a second group of solutions and a $+360°k$ term. But there is good news: we can ignore the $+360°k$ term. This is because we know the angle must be between zero and $180°$ because it is in a triangle. However, we cannot ignore the second solution: we must figure out if it gives us a real answer or not. (Remember that the second solution for the sine function comes from $180^{{ \circ }} - \theta$ , where $\theta$ is whatever angle you get from your calculator.)
$\begin{aligned} x & = \sin^{-1}\left( \text{0.81399...} \right) \\x & = \text{54.48811...}° \quad \text{or} \quad \text{125.51188...}° \end{aligned}$

We have two answers and we need to figure out which of these answers is correct. Or are they both correct? It turns out that the second answer, $x \approx \text{125.5}°$ , is impossible. We know this by comparing to the diagram: the angle we want is an acute angle (less than $\text{90}\text{°}$ ). So while $\text{125.5}°$ solves the equation, it is not a valid angle for the triangle in this question.

NOTE: It is possible that a question like this will have two solutions, or that the obtuse angle is the only correct answer. For example, if the diagram was not drawn to scale, we could not assume that the angle is acute. You should not assume that the answer will always be an acute angle!

Rounded to the tenths place, the angle has a value of $x = \text{54.5}°$ .

Submit your answer as:

ID is: 3500 Seed is: 3449

The flexibility of the sine rule

The diagram below shows a triangle, drawn to scale. Two angles are shown, with values of $37°$ and $103°$ . The side of the triangle across from the $103^{ \circ }$ angle has a length of $\text{9.8}$ units.

Suppose you need to calculate $x$ , the side across from the $37^{ \circ }$ angle. The two equations below are both accurate and you can use either of these versions to find $x$ . But one choice makes the solution a bit easier. Which version is the better choice?

Version 1	$\dfrac{\sin \left(\text{37}°\right)}{x} = \dfrac{\sin \left(\text{103}° \right)}{ \text{9.8} }$
Version 2	$\dfrac{x}{\sin \left(\text{37}°\right)} = \dfrac{ \text{9.8} }{\sin \left(\text{103}° \right)}$

Answer: The better choice is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about solving equations these two equations for $y$ :

\begin{matrix} \frac{y}{5} = \frac{7}{10} or \frac{5}{y} = \frac{10}{7} \end{matrix}

$\begin{array}{c}\frac{y}{5} = \frac{7}{10} \quad \text{or} \quad \frac{5}{y} = \frac{10}{7}\end{array}$

The equation on the left is more convenient to solve. Why?

STEP: Choose the version with the needed value in the numerator
[−1 point ⇒ 0 / 1 points left]

We can arrange the sine rule in two different ways. We can either put the angles in the numerator with the sides in the denominator, or the other way around:

\begin{aligned} _{sides in denominators:}^{Angles in numerators and} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ _{angles in denominators:}^{Sides in numerators and} \frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} ^{\text{Angles in numerators and}}_{\;\text{sides in denominators:}} \quad\frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\[0.4cm] ^{\text{Sides in numerators and}}_{\;\text{angles in denominators:}} \quad\frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}$

In both equations, $a$ and $b$ represent sides of the triangle and $\hat{A}$ and $\hat{B}$ represent angles. Solving equations with the unknown value in the numerator is usually easier than solving equations with the unknown in the denominator. So we should use the arrangement which will put the value we want to find in the numerator.

In this question, we have two versions of the sine rule for the triangle given. Both of these versions are correct. But since we want to find a side the choice with the sides in the numerators is the better choice.

The correct choice is Version 2.

Submit your answer as:

Now solve for the missing side, $x$ . Use the equation identified in Question 1.

INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.

Answer: The length of the missing side is units.
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start with the equation we identified in Question 1:

$\begin{array}{c}\dfrac{x}{\sin \left(\text{37}°\right)} = \dfrac{ \text{9.8} }{\sin \left(\text{103}° \right)}\end{array}$

You need to solve for $x$ .

STEP: Solve for the missing side
[−2 points ⇒ 0 / 2 points left]

We need to find the value of $x$ , which is the missing side. We already have the equation from Question 1. So let's solve it! Start by evaluating the right side of the equation.

$\begin{aligned} \frac{x}{\sin \left(\text{37}°\right)} & = \frac{ \text{9.8} }{\sin \left(\text{103}° \right)} \\ \frac{x}{\sin \left(\text{37}°\right)} & = \frac{ \text{9.8} }{ \left( \text{0.97437...} \right)} \\\frac{x}{\sin \left(\text{37}°\right)} & = \text{10.05778...} \end{aligned}$
Now work on the value of the fraction on the right side and multiply both sides of the equation by the $\sin \left(\text{37}° \right)$ to cancel the denominator on the left.
$\begin{aligned} x & = \left( \text{10.05778...} \right) \cdot \sin \left(\text{37}° \right) \\x & = \text{6.05292...} \\x & \approx \text{6.1} \end{aligned}$

And that's the answer. The final step rounds the answer to one decimal place, as required in the instructions.

The unknown side of the triangle has a length of $\text{6.1}$ .

Submit your answer as:

ID is: 3500 Seed is: 2126

The flexibility of the sine rule

The diagram below shows a triangle, drawn to scale. One of the sides has a length of $\text{8.2}$ and another side has the length $\text{14.7}$ . The angle across from the side labelled $\text{14.7}$ is $\text{91}°$ .

Suppose you need to calculate $x$ , the angle across from the side with length $\text{8.2}$ . The two equations below are both accurate and you can use either of these versions to find $x$ . But one choice makes the solution a bit easier. Which version is the better choice?

Version 1	$\dfrac{\sin x}{\text{8.2}} = \dfrac{\sin \left( \text{91}° \right)}{ \text{14.7}}$
Version 2	$\dfrac{\text{8.2}}{\sin x} = \dfrac{ \text{14.7}}{\sin \left( \text{91}° \right)}$

Answer: The better choice is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Think about solving equations these two equations for $y$ :

\begin{matrix} \frac{y}{5} = \frac{7}{10} or \frac{5}{y} = \frac{10}{7} \end{matrix}

$\begin{array}{c}\frac{y}{5} = \frac{7}{10} \quad \text{or} \quad \frac{5}{y} = \frac{10}{7}\end{array}$

The equation on the left is more convenient to solve. Why?

STEP: Choose the version with the needed value in the numerator
[−1 point ⇒ 0 / 1 points left]

We can arrange the sine rule in two different ways. We can either put the angles in the numerator with the sides in the denominator, or the other way around:

\begin{aligned} _{sides in denominators:}^{Angles in numerators and} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ _{angles in denominators:}^{Sides in numerators and} \frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} ^{\text{Angles in numerators and}}_{\;\text{sides in denominators:}} \quad\frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\[0.4cm] ^{\text{Sides in numerators and}}_{\;\text{angles in denominators:}} \quad\frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}$

In both equations, $a$ and $b$ represent sides of the triangle and $\hat{A}$ and $\hat{B}$ represent angles. Solving equations with the unknown value in the numerator is usually easier than solving equations with the unknown in the denominator. So we should use the arrangement which will put the value we want to find in the numerator.

In this question, we have two versions of the sine rule for the triangle given. Both of these versions are correct. But since we want to find an angle the choice with the angles in the numerators is the better choice.

The correct choice is Version 1.

Submit your answer as:

Now solve for the missing angle, $x$ . Use the equation identified in Question 1.

INSTRUCTION: Round your answer to the tenths place (one decimal) if necessary.

Answer: The missing angle is °
numeric
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start with the equation we identified in Question 1:

$\begin{array}{c}\dfrac{\sin x}{\text{8.2}} = \dfrac{\sin \left( \text{91}° \right)}{ \text{14.7}}\end{array}$

You need to solve for $x$ .

STEP: Solve for the missing angle
[−3 points ⇒ 0 / 3 points left]

We need to find the value of $x$ , which is the missing angle. We already have the equation from Question 1. So let's solve it! Start by evaluating the right side of the equation.

$\begin{aligned} \frac{\sin x}{\text{8.2}} & = \frac{\sin \left( \text{91}° \right)}{ \text{14.7}} \\ \frac{\sin x}{\text{8.2}} & = \frac{ \left( \text{0.99984...} \right) }{ \text{14.7}} \\\frac{\sin x}{\text{8.2}} & = \text{0.06801...} \\\sin x & = \left( \text{0.06801...} \right)\left( \text{8.2} \right) \\\sin x & = \text{0.55773...} \end{aligned}$
Now we must use the inverse sine operation. This comes with a second group of solutions and a $+360°k$ term. But there is good news: we can ignore the $+360°k$ term. This is because we know the angle must be between zero and $180°$ because it is in a triangle. However, we cannot ignore the second solution: we must figure out if it gives us a real answer or not. (Remember that the second solution for the sine function comes from $180^{{ \circ }} - \theta$ , where $\theta$ is whatever angle you get from your calculator.)
$\begin{aligned} x & = \sin^{-1}\left( \text{0.55773...} \right) \\x & = \text{33.89952...}° \quad \text{or} \quad \text{146.10047...}° \end{aligned}$

We have two answers and we need to figure out which of these answers is correct. Or are they both correct? It turns out that the second answer, $x \approx \text{146.1}°$ , is impossible. We know this by comparing to the diagram: the angle we want is an acute angle (less than $\text{90}\text{°}$ ). So while $\text{146.1}°$ solves the equation, it is not a valid angle for the triangle in this question. (In this case, there is another problem with $\text{146.1}°$ : if we add it to the angle given in the question, we get $\text{146.1}° + 91° = \text{237.1}°$ . And the angles in a triangle cannot be larger than $\text{180}\text{°}$ .)

NOTE: It is possible that a question like this will have two solutions, or that the obtuse angle is the only correct answer. For example, if the diagram was not drawn to scale, we could not assume that the angle is acute. You should not assume that the answer will always be an acute angle!

Rounded to the tenths place, the angle has a value of $x = \text{33.9}°$ .

Submit your answer as:

ID is: 1433 Seed is: 6722

The sine rule: identifying the correct sides and angles

The triangle below has sides labelled $x, y$ , and $z$ , and angles labelled $X, Y$ , and $Z$ .

For this triangle, the sine rule states:

\begin{aligned} \frac{?}{\sin X} = \frac{y}{\sin Y} \end{aligned}

$\begin{aligned} \frac{\; ? \;}{\sin X} = \frac{y}{\sin Y} \end{aligned}$

Which of the following choices belongs in the place of the '?' in the formula?

Answer: The missing symbol is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule has an angle and a side. The side must be the side opposite from the angle. Look in the picture of the triangle for sides and angles which are across from (opposite) each other in the triangle.

STEP: Connect the sides and angles in the diagram according to the sine rule
[−1 point ⇒ 0 / 1 points left]

In the sine rule, each fraction must contain a side and an angle. These values (the side and the angle) must be opposite (across from) each other in the triangle. For example, in the triangle in this question, the angle $X$ and side $x$ are opposite each other.

We are given the formula:

\begin{aligned} \frac{?}{\sin X} = \frac{y}{\sin Y} \end{aligned}

$\begin{aligned} \frac{\; ? \;}{\sin X} = \frac{y}{\sin Y} \end{aligned}$

The question mark is in the position of one of the sides. The fraction with the question mark includes the angle $X$ . The side that we need must be opposite from that angle. Looking at the diagram, we can see that the answer is side $x$ .

The missing variable in the equation is the side x.

Submit your answer as:

ID is: 1433 Seed is: 6135

The sine rule: identifying the correct sides and angles

The triangle below has sides labelled $c, b$ , and $a$ , and angles labelled $C, B$ , and $A$ .

For this triangle, the sine rule states:

\begin{aligned} \frac{b}{\sin ?} = \frac{c}{\sin C} \end{aligned}

$\begin{aligned} \frac{b}{\sin \; ? \;} = \frac{c}{\sin C} \end{aligned}$

Which of the following choices belongs in the place of the '?' in the formula?

Answer: The missing symbol is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule has an angle and a side. The side must be the side opposite from the angle. Look in the picture of the triangle for sides and angles which are across from (opposite) each other in the triangle.

STEP: Connect the sides and angles in the diagram according to the sine rule
[−1 point ⇒ 0 / 1 points left]

In the sine rule, each fraction must contain a side and an angle. These values (the side and the angle) must be opposite (across from) each other in the triangle. For example, in the triangle in this question, the angle $B$ and side $b$ are opposite each other.

We are given the formula:

\begin{aligned} \frac{b}{\sin ?} = \frac{c}{\sin C} \end{aligned}

$\begin{aligned} \frac{b}{\sin \; ? \;} = \frac{c}{\sin C} \end{aligned}$

The question mark is in the position of one of the angles. Since the fraction with the question mark has the side $b$ in it, the angle we need must be opposite from side $b$ .

The missing variable in the equation is the angle B.

Submit your answer as:

ID is: 1433 Seed is: 6235

The sine rule: identifying the correct sides and angles

The triangle below has sides labelled $a, m$ , and $b$ , and angles labelled $\alpha, \theta$ , and $\beta$ .

For this triangle, the sine rule states:

\begin{aligned} \frac{m}{\sin θ} = \frac{?}{\sin β} \end{aligned}

$\begin{aligned} \frac{m}{\sin \theta} = \frac{\; ? \;}{\sin \beta} \end{aligned}$

Which of the following choices belongs in the place of the '?' in the formula?

Answer: The missing symbol is: .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Each fraction in the sine rule has an angle and a side. The side must be the side opposite from the angle. Look in the picture of the triangle for sides and angles which are across from (opposite) each other in the triangle.

STEP: Connect the sides and angles in the diagram according to the sine rule
[−1 point ⇒ 0 / 1 points left]

In the sine rule, each fraction must contain a side and an angle. These values (the side and the angle) must be opposite (across from) each other in the triangle. For example, in the triangle in this question, the angle $\theta$ and side $m$ are opposite each other.

We are given the formula:

\begin{aligned} \frac{m}{\sin θ} = \frac{?}{\sin β} \end{aligned}

$\begin{aligned} \frac{m}{\sin \theta} = \frac{\; ? \;}{\sin \beta} \end{aligned}$

The question mark is in the position of one of the sides. The fraction with the question mark includes the angle $\beta$ . The side that we need must be opposite from that angle. Looking at the diagram, we can see that the answer is side $b$ .

The missing variable in the equation is the side b.

Submit your answer as:

ID is: 3559 Seed is: 9915

Find the missing length

In triangle $EDF$ , two angles are known: $\hat{E} = 44°$ and $\hat{D} = 49°$ . Sides $\overline{DF}$ and $\overline{FE}$ have lengths $5 w - 2$ and $10 w^{2} - 4 w$ respectively, where $w \in \mathbb{R}$ . All lengths are non-zero and have units of $\text{cm}$ .

Find then length of $\overline{DF}$ .

INSTRUCTION: Round your answer to two decimal places.

Answer: The length of

\bar{D F}

$\overline{DF}$ is

cm

$\text{cm}$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by drawing a diagram of the triangle. Label it with everything you know. Then use the sine rule to write an equation which includes the variable $w$ .

STEP: Draw a picture of the triangle and label it
[−1 point ⇒ 4 / 5 points left]

The first thing to do is draw a picture. The question tells us a lot about a triangle, but we need to organise that information in a diagram. It is important to label everything you know about the triangle.

NOTE: The diagram does not need to be to scale. But the labels should be accurate. And since the diagram is not to scale, we cannot draw any conclusions from the picture itself.

STEP: Write an equation including

w

$w$
[−1 point ⇒ 3 / 5 points left]

We need to determine the value of $w$ . We can do this using the sine rule. We can arrange the sine rule in various ways. In this case, the expression $10 w^{2} - 4 w$ can be factorised, and it has a common factor with $5 w - 2$ ! So it looks like we will be able to cancel something. Let's put the side equal to $10 w^{2} - 4 w$ in the left numerator so that we can get a more convenient equation to solve.

\begin{matrix} \frac{10 w^{2} - 4 w}{\sin 49 °} = \frac{5 w - 2}{\sin 44 °} \end{matrix}

$\begin{array}{c} \dfrac{10 w^{2} - 4 w}{\sin 49°} = \dfrac{5 w - 2}{\sin 44°} \end{array}$

STEP: Solve for

w

$w$
[−2 points ⇒ 1 / 5 points left]

We need to get the $w$ terms on one side of the equation. After that, we can factorise and simplify on the left side.

\begin{aligned} \frac{10 w^{2} - 4 w}{\sin 49 °} & = \frac{5 w - 2}{\sin 44 °} \\ \frac{10 w^{2} - 4 w}{5 w - 2} & = \frac{\sin 49 °}{\sin 44 °} \\ \frac{(2 w) (5 w - 2)}{5 w - 2} & = \frac{\sin 49 °}{\sin 44 °} \\ 2 w & = \frac{\sin 49 °}{\sin 44 °} \end{aligned}

$\begin{aligned} \frac{10 w^{2} - 4 w}{\sin 49°} & = \frac{5 w - 2}{\sin 44°} \\[0.15cm] \frac{10 w^{2} - 4 w}{5 w - 2} & = \frac{\sin 49°}{\sin 44°} \\[0.15cm] \frac{\left(2 w\right)\left(5 w - 2\right)}{5 w - 2} & = \frac{\sin 49°}{\sin 44°} \\[0.15cm] 2 w & = \frac{\sin 49°}{\sin 44°} \\[0.15cm] \end{aligned}$

Now isolate the $w$ on the left side. You can evaluate the sine functions on the right now if you want, but leaving them will be helpful because we can avoid dragging long decimal numbers around.

\begin{aligned} 2 w & = \frac{\sin 49 °}{\sin 44 °} \\ w & = (\frac{\sin 49 °}{\sin 44 °}) \cdot \frac{1}{2} \\ w & = 0.54322... \end{aligned}

$\begin{aligned} 2 w & = \frac{\sin 49°}{\sin 44°} \\ w & = \left(\frac{\sin 49°} {\sin 44°}\right) \cdot \frac{1}{2} \\ w & = \text{0.54322...} \end{aligned}$

Do not round this number yet - we need to use it in the final calculation.

STEP: Compute the length of

\bar{D F}

$\overline{DF}$
[−1 point ⇒ 0 / 5 points left]

Finally we can evaluate the length of $\overline{DF}$ .

\begin{aligned} \bar{D F} & = 5 w - 2 \\ = 5 (0.54322...) - 2 \\ = 0.71611... \\ \approx 0.72 cm \end{aligned}

$\begin{aligned} \overline{DF} & = 5 w - 2 \\ & = 5 \left(\text{0.54322...}\right) - 2 \\ & = \text{0.71611...} \\ & \approx \text{0.72}\ \text{cm} \end{aligned}$

As long as this number is positive, it is acceptable (lengths cannot be negative). So we are done.

The length of $\overline{DF}$ is $\text{0.72}\ \text{cm}$ .

Submit your answer as:

ID is: 3559 Seed is: 913

Find the missing length

In triangle $ACB$ , two angles are known: $\hat{A} = 78°$ and $\hat{C} = 70°$ . Sides $\overline{CB}$ and $\overline{BA}$ have lengths $3 w + 2$ and $9 w^{2} - 4$ respectively, where $w \in \mathbb{R}$ . All lengths are non-zero and have units of $\text{cm}$ .

Determine then length of $\overline{BA}$ .

INSTRUCTION: Round your answer to two decimal places.

Answer: The length of

\bar{B A}

$\overline{BA}$ is

cm

$\text{cm}$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by drawing a diagram of the triangle. Label it with everything you know. Then use the sine rule to write an equation which includes the variable $w$ .

STEP: Draw a picture of the triangle and label it
[−1 point ⇒ 4 / 5 points left]

The first thing to do is draw a picture. The question tells us a lot about a triangle, but we need to organise that information in a diagram. It is important to label everything you know about the triangle.

NOTE: The diagram does not need to be to scale. But the labels should be accurate. And since the diagram is not to scale, we cannot draw any conclusions from the picture itself.

STEP: Write an equation including

w

$w$
[−1 point ⇒ 3 / 5 points left]

We need to determine the value of $w$ . We can do this using the sine rule. We can arrange the sine rule in various ways. In this case, the expression $9 w^{2} - 4$ can be factorised, and it has a common factor with $3 w + 2$ ! So it looks like we will be able to cancel something. Let's put the side equal to $9 w^{2} - 4$ in the left numerator so that we can get a more convenient equation to solve.

\begin{matrix} \frac{9 w^{2} - 4}{\sin 70 °} = \frac{3 w + 2}{\sin 78 °} \end{matrix}

$\begin{array}{c} \dfrac{9 w^{2} - 4}{\sin 70°} = \dfrac{3 w + 2}{\sin 78°} \end{array}$

STEP: Solve for

w

$w$
[−2 points ⇒ 1 / 5 points left]

We need to get the $w$ terms on one side of the equation. After that, we can factorise and simplify on the left side.

\begin{aligned} \frac{9 w^{2} - 4}{\sin 70 °} & = \frac{3 w + 2}{\sin 78 °} \\ \frac{9 w^{2} - 4}{3 w + 2} & = \frac{\sin 70 °}{\sin 78 °} \\ \frac{(3 w - 2) (3 w + 2)}{3 w + 2} & = \frac{\sin 70 °}{\sin 78 °} \\ 3 w - 2 & = \frac{\sin 70 °}{\sin 78 °} \end{aligned}

$\begin{aligned} \frac{9 w^{2} - 4}{\sin 70°} & = \frac{3 w + 2}{\sin 78°} \\[0.15cm] \frac{9 w^{2} - 4}{3 w + 2} & = \frac{\sin 70°}{\sin 78°} \\[0.15cm] \frac{\left(3 w - 2\right)\left(3 w + 2\right)}{3 w + 2} & = \frac{\sin 70°}{\sin 78°} \\[0.15cm] 3 w - 2 & = \frac{\sin 70°}{\sin 78°} \\[0.15cm] \end{aligned}$

Now isolate the $w$ on the left side. You can evaluate the sine functions on the right now if you want, but leaving them will be helpful because we can avoid dragging long decimal numbers around.

\begin{aligned} 3 w - 2 & = \frac{\sin 70 °}{\sin 78 °} \\ w & = (\frac{\sin 70 °}{\sin 78 °} + 2) \cdot \frac{1}{3} \\ w & = 0.98689... \end{aligned}

$\begin{aligned} 3 w - 2 & = \frac{\sin 70°}{\sin 78°} \\ w & = \left(\frac{\sin 70°} {\sin 78°} +2 \right) \cdot \frac{1}{3} \\ w & = \text{0.98689...} \end{aligned}$

Do not round this number yet - we need to use it in the final calculation.

STEP: Compute the length of

\bar{B A}

$\overline{BA}$
[−1 point ⇒ 0 / 5 points left]

Finally we can evaluate the length of $\overline{BA}$ .

\begin{aligned} \bar{B A} & = 9 w^{2} - 4 \\ = 9 {(0.98689...)}^{2} - 4 \\ = 4.76566... \\ \approx 4.77 cm \end{aligned}

$\begin{aligned} \overline{BA} & = 9 w^{2} - 4 \\ & = 9 \left(\text{0.98689...}\right)^{2} - 4 \\ & = \text{4.76566...} \\ & \approx \text{4.77}\ \text{cm} \end{aligned}$

As long as this number is positive, it is acceptable (lengths cannot be negative). So we are done.

The length of $\overline{BA}$ is $\text{4.77}\ \text{cm}$ .

Submit your answer as:

ID is: 3559 Seed is: 6533

Find the missing length

In triangle $EDF$ , two angles are known: $\hat{E} = 69°$ and $\hat{D} = 50°$ . Sides $\overline{DF}$ and $\overline{FE}$ have lengths $9 d^{2} - 4$ and $3 d + 2$ respectively, where $d \in \mathbb{R}$ . All lengths are non-zero and have units of $\text{mm}$ .

Determine then length of $\overline{FE}$ .

INSTRUCTION: Round your answer to two decimal places.

Answer: The length of

\bar{F E}

$\overline{FE}$ is

mm

$\text{mm}$ .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 5 / 5 points left]

Start by drawing a diagram of the triangle. Label it with everything you know. Then use the sine rule to write an equation which includes the variable $d$ .

STEP: Draw a picture of the triangle and label it
[−1 point ⇒ 4 / 5 points left]

The first thing to do is draw a picture. The question tells us a lot about a triangle, but we need to organise that information in a diagram. It is important to label everything you know about the triangle.

NOTE: The diagram does not need to be to scale. But the labels should be accurate. And since the diagram is not to scale, we cannot draw any conclusions from the picture itself.

STEP: Write an equation including

d

$d$
[−1 point ⇒ 3 / 5 points left]

We need to determine the value of $d$ . We can do this using the sine rule. We can arrange the sine rule in various ways. In this case, the expression $9 d^{2} - 4$ can be factorised, and it has a common factor with $3 d + 2$ ! So it looks like we will be able to cancel something. Let's put the side equal to $9 d^{2} - 4$ in the left numerator so that we can get a more convenient equation to solve.

\begin{matrix} \frac{9 d^{2} - 4}{\sin 69 °} = \frac{3 d + 2}{\sin 50 °} \end{matrix}

$\begin{array}{c} \dfrac{9 d^{2} - 4}{\sin 69°} = \dfrac{3 d + 2}{\sin 50°} \end{array}$

STEP: Solve for

d

$d$
[−2 points ⇒ 1 / 5 points left]

We need to get the $d$ terms on one side of the equation. After that, we can factorise and simplify on the left side.

\begin{aligned} \frac{9 d^{2} - 4}{\sin 69 °} & = \frac{3 d + 2}{\sin 50 °} \\ \frac{9 d^{2} - 4}{3 d + 2} & = \frac{\sin 69 °}{\sin 50 °} \\ \frac{(3 d - 2) (3 d + 2)}{3 d + 2} & = \frac{\sin 69 °}{\sin 50 °} \\ 3 d - 2 & = \frac{\sin 69 °}{\sin 50 °} \end{aligned}

$\begin{aligned} \frac{9 d^{2} - 4}{\sin 69°} & = \frac{3 d + 2}{\sin 50°} \\[0.15cm] \frac{9 d^{2} - 4}{3 d + 2} & = \frac{\sin 69°}{\sin 50°} \\[0.15cm] \frac{\left(3 d - 2\right)\left(3 d + 2\right)}{3 d + 2} & = \frac{\sin 69°}{\sin 50°} \\[0.15cm] 3 d - 2 & = \frac{\sin 69°}{\sin 50°} \\[0.15cm] \end{aligned}$

Now isolate the $d$ on the left side. You can evaluate the sine functions on the right now if you want, but leaving them will be helpful because we can avoid dragging long decimal numbers around.

\begin{aligned} 3 d - 2 & = \frac{\sin 69 °}{\sin 50 °} \\ d & = (\frac{\sin 69 °}{\sin 50 °} + 2) \cdot \frac{1}{3} \\ d & = 1.07290... \end{aligned}

$\begin{aligned} 3 d - 2 & = \frac{\sin 69°}{\sin 50°} \\ d & = \left(\frac{\sin 69°} {\sin 50°} +2 \right) \cdot \frac{1}{3} \\ d & = \text{1.07290...} \end{aligned}$

Do not round this number yet - we need to use it in the final calculation.

STEP: Compute the length of

\bar{F E}

$\overline{FE}$
[−1 point ⇒ 0 / 5 points left]

Finally we can evaluate the length of $\overline{FE}$ .

\begin{aligned} \bar{F E} & = 3 d + 2 \\ = 3 (1.07290...) + 2 \\ = 5.21870... \\ \approx 5.22 mm \end{aligned}

$\begin{aligned} \overline{FE} & = 3 d + 2 \\ & = 3 \left(\text{1.07290...}\right) + 2 \\ & = \text{5.21870...} \\ & \approx \text{5.22}\ \text{mm} \end{aligned}$

As long as this number is positive, it is acceptable (lengths cannot be negative). So we are done.

The length of $\overline{FE}$ is $\text{5.22}\ \text{mm}$ .

Submit your answer as:

ID is: 3581 Seed is: 7381

Connecting the sine rule to right triangles

The sine rule applies to any triangle. This includes right-angled triangles. Which statement below identifies what happens to the sine rule if one of the angles is $\text{90}\text{°}$ ?

Choices	If an angle is $\text{90}\text{°}$ , the sine rule...
A	only works for the acute angles.
B	will give two answers due to the ambiguous case.
C	becomes the same as the theorom of Pythagoras.
D	becomes the sine ratio (opposite & hypotenuse).

Answer:

The correct statement is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Draw a triangle to represent the situation described in the question. Then work out the sine rule including the right angle.

STEP: Find the link between the sine rule and right-angled triangles
[−1 point ⇒ 0 / 1 points left]

This question is about what happens when we use a $\text{90}\text{°}$ angle in the sine rule. If one of the angles is $\text{90}\text{°}$ then we have a right-angled triangle. For example:

Since this is a right-angled triangle, the normal trigonometric ratios (sine, cosine, and tangent) apply to the triangle. But so does the sine rule. And if we include the right angle when we write the sine rule, it simplifies into the sine ratio for right-angled triangles. (Question 2 is about why this happens.)

The correct choice is D.

Submit your answer as:

The table below shows (proves) that if we use the sine rule with a $\text{90}\text{°}$ angle it simplifies into the sine ratio for right-angled triangles. However, there is one missing step and one missing reason. Complete the proof by choosing the correct options from the lists available.

Answer:

Steps	Reasons
	Given
$\dfrac{\sin X}{x} = \dfrac{\sin Z}{z}$	Sine rule for $\Delta XYZ$
	Substitute $Z = 90°$
$\dfrac{\sin X}{x} = \dfrac{1}{z}$
$\sin X = \dfrac{x}{z}$	Multiply by $x$
$\sin X = \dfrac{\text{opposite}}{\text{hypotenuse}}$	Definition of opposite & hypotenuse

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by analysing the steps and reasons which are already shown. Think about what they mean. Then try to decide what the missing pieces should be.

STEP: Select the correct step and reason
[−2 points ⇒ 0 / 2 points left]

The proof in this question connects the sine rule to right-angled triangles. In such a case, the sine rule becomes the sine ratio for a right-angled triangle. In other words, the sine rule ends up being equivalent to the ratio "opposite over hypotenuse." This connection between the sine rule and the sine ratio exists because $\sin\text{90}\text{°}$ is equal to $\text{1}$ .

The correct answers in the proof are shown below.

Steps	Reasons
	Given
$\dfrac{\sin X}{x} = \dfrac{\sin Z}{z}$	Sine rule for $\Delta XYZ$
$\dfrac{\sin X}{x} = \dfrac{\sin (90°)}{z}$	Substitute $Z = 90°$
$\dfrac{\sin X}{x} = \dfrac{1}{z}$	Evaluate $\sin 90° = 1$
$\sin X = \dfrac{x}{z}$	Multiply by $x$
$\sin X = \dfrac{\text{opposite}}{\text{hypotenuse}}$	Definition of opposite & hypotenuse

Submit your answer as: and

ID is: 3581 Seed is: 1269

Connecting the sine rule to right triangles

The sine rule applies to any triangle. This includes right-angled triangles. Which statement below identifies what happens to the sine rule if one of the angles is $\text{90}\text{°}$ ?

Choices	If an angle is $\text{90}\text{°}$ , the sine rule...
A	becomes the sine ratio (opposite & hypotenuse).
B	only works for the acute angles.
C	becomes the same as the theorom of Pythagoras.
D	only uses the reciprocals.

Answer:

The correct statement is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Draw a triangle to represent the situation described in the question. Then work out the sine rule including the right angle.

STEP: Find the link between the sine rule and right-angled triangles
[−1 point ⇒ 0 / 1 points left]

This question is about what happens when we use a $\text{90}\text{°}$ angle in the sine rule. If one of the angles is $\text{90}\text{°}$ then we have a right-angled triangle. For example:

Since this is a right-angled triangle, the normal trigonometric ratios (sine, cosine, and tangent) apply to the triangle. But so does the sine rule. And if we include the right angle when we write the sine rule, it simplifies into the sine ratio for right-angled triangles. (Question 2 is about why this happens.)

The correct choice is A.

Submit your answer as:

The table below shows (proves) that if we use the sine rule with a $\text{90}\text{°}$ angle it simplifies into the sine ratio for right-angled triangles. However, there is one missing step and one missing reason. Complete the proof by choosing the correct options from the lists available.

Answer:

Steps	Reasons
	Given
$\dfrac{\sin X}{x} = \dfrac{\sin Z}{z}$
$\dfrac{\sin X}{x} = \dfrac{\sin (90°)}{z}$	Substitute $Z = 90°$
	Evaluate $\sin 90° = 1$
$\sin X = \dfrac{x}{z}$	Multiply by $x$
$\sin X = \dfrac{\text{opposite}}{\text{hypotenuse}}$	Definition of opposite & hypotenuse

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by analysing the steps and reasons which are already shown. Think about what they mean. Then try to decide what the missing pieces should be.

STEP: Select the correct step and reason
[−2 points ⇒ 0 / 2 points left]

The proof in this question connects the sine rule to right-angled triangles. In such a case, the sine rule becomes the sine ratio for a right-angled triangle. In other words, the sine rule ends up being equivalent to the ratio "opposite over hypotenuse." This connection between the sine rule and the sine ratio exists because $\sin\text{90}\text{°}$ is equal to $\text{1}$ .

The correct answers in the proof are shown below.

Steps	Reasons
	Given
$\dfrac{\sin X}{x} = \dfrac{\sin Z}{z}$	Sine rule for $\Delta XYZ$
$\dfrac{\sin X}{x} = \dfrac{\sin (90°)}{z}$	Substitute $Z = 90°$
$\dfrac{\sin X}{x} = \dfrac{1}{z}$	Evaluate $\sin 90° = 1$
$\sin X = \dfrac{x}{z}$	Multiply by $x$
$\sin X = \dfrac{\text{opposite}}{\text{hypotenuse}}$	Definition of opposite & hypotenuse

Submit your answer as: and

ID is: 3581 Seed is: 6701

Connecting the sine rule to right triangles

The sine rule applies to any triangle. This includes right-angled triangles. Which statement below identifies what happens to the sine rule if one of the angles is $\text{90}\text{°}$ ?

Choices	If an angle is $\text{90}\text{°}$ , the sine rule...
A	becomes the sine ratio (opposite & hypotenuse).
B	will not work anymore.
C	will give two answers due to the ambiguous case.
D	becomes the same as the theorom of Pythagoras.

Answer:

The correct statement is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Draw a triangle to represent the situation described in the question. Then work out the sine rule including the right angle.

STEP: Find the link between the sine rule and right-angled triangles
[−1 point ⇒ 0 / 1 points left]

This question is about what happens when we use a $\text{90}\text{°}$ angle in the sine rule. If one of the angles is $\text{90}\text{°}$ then we have a right-angled triangle. For example:

Since this is a right-angled triangle, the normal trigonometric ratios (sine, cosine, and tangent) apply to the triangle. But so does the sine rule. And if we include the right angle when we write the sine rule, it simplifies into the sine ratio for right-angled triangles. (Question 2 is about why this happens.)

The correct choice is A.

Submit your answer as:

The table below shows (proves) that if we use the sine rule with a $\text{90}\text{°}$ angle it simplifies into the sine ratio for right-angled triangles. However, there is one missing step and one missing reason. Complete the proof by choosing the correct options from the lists available.

Answer:

Steps	Reasons
	Given
	Sine rule for $\Delta XYZ$
$\dfrac{\sin X}{x} = \dfrac{\sin (90°)}{z}$
$\dfrac{\sin X}{x} = \dfrac{1}{z}$	Evaluate $\sin 90° = 1$
$\sin X = \dfrac{x}{z}$	Multiply by $x$
$\sin X = \dfrac{\text{opposite}}{\text{hypotenuse}}$	Definition of opposite & hypotenuse

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 2 / 2 points left]

Start by analysing the steps and reasons which are already shown. Think about what they mean. Then try to decide what the missing pieces should be.

STEP: Select the correct step and reason
[−2 points ⇒ 0 / 2 points left]

The proof in this question connects the sine rule to right-angled triangles. In such a case, the sine rule becomes the sine ratio for a right-angled triangle. In other words, the sine rule ends up being equivalent to the ratio "opposite over hypotenuse." This connection between the sine rule and the sine ratio exists because $\sin\text{90}\text{°}$ is equal to $\text{1}$ .

The correct answers in the proof are shown below.

Steps	Reasons
	Given
$\dfrac{\sin X}{x} = \dfrac{\sin Z}{z}$	Sine rule for $\Delta XYZ$
$\dfrac{\sin X}{x} = \dfrac{\sin (90°)}{z}$	Substitute $Z = 90°$
$\dfrac{\sin X}{x} = \dfrac{1}{z}$	Evaluate $\sin 90° = 1$
$\sin X = \dfrac{x}{z}$	Multiply by $x$
$\sin X = \dfrac{\text{opposite}}{\text{hypotenuse}}$	Definition of opposite & hypotenuse

Submit your answer as: and

ID is: 3571 Seed is: 8200

The flexibility of the sine rule

The sine rule describes the relationship between the sides and angles in any triangle. The ratio of any side to the sine of the angle opposite from that side is the same for all three side-angle pairs in the triangle. For example, consider this triangle, which has sides $m$ , $p$ , and $n$ , and angles $R$ , $S$ , and $Q$ . Two of the three equal ratios are shown below the triangle.

\begin{matrix} \frac{\sin S}{n} = \frac{\sin Q}{m} = third ratio \end{matrix}

$\begin{array}{c} \dfrac{ \sin S }{ n } = \dfrac{ \sin Q }{ m } = \text{third ratio} \end{array}$

Which of the choices below is the third equal ratio for this triangle?

A	$\dfrac{ \sin R }{ n }$
B	$\dfrac{ \sin R }{ p }$
C	$\dfrac{ \sin S }{ p }$

Answer: The third ratio is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by figuring out which side and which angle are not used in the two ratios given in the question. They must be in the answer.

STEP: Compare the given ratios to the triangle
[−1 point ⇒ 0 / 1 points left]

This question is about how we can apply the sine rule to a triangle. Each ratio (fraction) in the sine rule contains one side and one angle in the triangle. For any triangle there are three equal ratios. In this question we have two of those ratios for the triangle shown above. We need to find the third equal ratio.

The sine rule always includes side-angle pairs which are across from each other in the triangle. There are always three of these pairs, as shown below:

Each of the linked pairs above corresponds to a ratio for the sine rule. In fact, the two ratios given in the question each include one of the linked pairs:

\begin{array}{ccc} \frac{\sin S}{n} & = & \frac{\sin Q}{m} \\ ↑ & ↑ \\ side n is & side m is \\ across from & across from \\ angle S & angle Q \end{array}

$\begin{array}{ccc} \dfrac{ \sin S }{ n } & = & \dfrac{ \sin Q }{ m } \\ \uparrow & & \uparrow \\ \text{side } n \text{ is} & & \text{side } m \text{ is} \\ \text{across from} & & \text{across from} \\ \text{angle } S & & \text{angle } Q \end{array}$

The third side-angle pair is side $p$ and angle $R$ . These values must be in the third ratio!

\begin{matrix} \frac{\sin S}{n} = \frac{\sin Q}{m} = \frac{\sin R}{p} \end{matrix}

$\begin{array}{c} \dfrac{ \sin S }{ n } = \dfrac{ \sin Q }{ m } = \dfrac{ \sin R }{ p } \end{array}$

NOTE:

We can also write the ratios upside down, but if we do this then all the ratios should match:

\begin{matrix} \frac{n}{\sin S} = \frac{m}{\sin Q} = \frac{p}{\sin R} \end{matrix}

$\begin{array}{c} \dfrac{ n }{ \sin S } = \dfrac{ m }{ \sin Q } = \dfrac{ p }{ \sin R } \end{array}$

The correct answer is $\dfrac{ \sin R }{ p }$ , which is choice B.

Submit your answer as:

ID is: 3571 Seed is: 1042

The flexibility of the sine rule

The sine rule describes the relationship between the sides and angles in any triangle. The ratio of any side to the sine of the angle opposite from that side is the same for all three side-angle pairs in the triangle. For example, consider this triangle, which has sides $p$ , $n$ , and $m$ , and angles $H$ , $G$ , and $F$ . Two of the three equal ratios are shown below the triangle.

\begin{matrix} \frac{m}{\sin G} = \frac{p}{\sin F} = third ratio \end{matrix}

$\begin{array}{c} \dfrac{ m }{ \sin G } = \dfrac{ p }{ \sin F } = \text{third ratio} \end{array}$

Which of the choices below is the third equal ratio for this triangle?

A	$\dfrac{ m }{ \sin H }$
B	$\dfrac{ n }{ \sin F }$
C	$\dfrac{ n }{ \sin H }$

Answer: The third ratio is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by figuring out which side and which angle are not used in the two ratios given in the question. They must be in the answer.

STEP: Compare the given ratios to the triangle
[−1 point ⇒ 0 / 1 points left]

This question is about how we can apply the sine rule to a triangle. Each ratio (fraction) in the sine rule contains one side and one angle in the triangle. For any triangle there are three equal ratios. In this question we have two of those ratios for the triangle shown above. We need to find the third equal ratio.

The sine rule always includes side-angle pairs which are across from each other in the triangle. There are always three of these pairs, as shown below:

Each of the linked pairs above corresponds to a ratio for the sine rule. In fact, the two ratios given in the question each include one of the linked pairs:

\begin{array}{ccc} \frac{m}{\sin G} & = & \frac{p}{\sin F} \\ ↑ & ↑ \\ side m is & side p is \\ across from & across from \\ angle G & angle F \end{array}

$\begin{array}{ccc} \dfrac{ m }{ \sin G } & = & \dfrac{ p }{ \sin F } \\ \uparrow & & \uparrow \\ \text{side } m \text{ is} & & \text{side } p \text{ is} \\ \text{across from} & & \text{across from} \\ \text{angle } G & & \text{angle } F \end{array}$

The third side-angle pair is side $n$ and angle $H$ . These values must be in the third ratio!

\begin{matrix} \frac{m}{\sin G} = \frac{p}{\sin F} = \frac{n}{\sin H} \end{matrix}

$\begin{array}{c} \dfrac{ m }{ \sin G } = \dfrac{ p }{ \sin F } = \dfrac{ n }{ \sin H } \end{array}$

NOTE:

We can also write the ratios upside down, but if we do this then all the ratios should match:

\begin{matrix} \frac{\sin G}{m} = \frac{\sin F}{p} = \frac{\sin H}{n} \end{matrix}

$\begin{array}{c} \dfrac{ \sin G }{ m } = \dfrac{ \sin F }{ p } = \dfrac{ \sin H }{ n } \end{array}$

The correct answer is $\dfrac{ n }{ \sin H }$ , which is choice C.

Submit your answer as:

ID is: 3571 Seed is: 8385

The flexibility of the sine rule

The sine rule describes the relationship between the sides and angles in any triangle. The ratio of any side to the sine of the angle opposite from that side is the same for all three side-angle pairs in the triangle. For example, consider this triangle, which has sides $m$ , $p$ , and $n$ , and angles $G$ , $H$ , and $F$ . Two of the three equal ratios are shown below the triangle.

\begin{matrix} \frac{n}{\sin H} = \frac{m}{\sin F} = third ratio \end{matrix}

$\begin{array}{c} \dfrac{ n }{ \sin H } = \dfrac{ m }{ \sin F } = \text{third ratio} \end{array}$

Which of the choices below is the third equal ratio for this triangle?

A	$\dfrac{ p }{ \sin H }$
B	$\dfrac{ p }{ \sin G }$
C	$\dfrac{ p }{ \sin F }$

Answer: The third ratio is choice .

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Start by figuring out which side and which angle are not used in the two ratios given in the question. They must be in the answer.

STEP: Compare the given ratios to the triangle
[−1 point ⇒ 0 / 1 points left]

This question is about how we can apply the sine rule to a triangle. Each ratio (fraction) in the sine rule contains one side and one angle in the triangle. For any triangle there are three equal ratios. In this question we have two of those ratios for the triangle shown above. We need to find the third equal ratio.

The sine rule always includes side-angle pairs which are across from each other in the triangle. There are always three of these pairs, as shown below:

Each of the linked pairs above corresponds to a ratio for the sine rule. In fact, the two ratios given in the question each include one of the linked pairs:

\begin{array}{ccc} \frac{n}{\sin H} & = & \frac{m}{\sin F} \\ ↑ & ↑ \\ side n is & side m is \\ across from & across from \\ angle H & angle F \end{array}

$\begin{array}{ccc} \dfrac{ n }{ \sin H } & = & \dfrac{ m }{ \sin F } \\ \uparrow & & \uparrow \\ \text{side } n \text{ is} & & \text{side } m \text{ is} \\ \text{across from} & & \text{across from} \\ \text{angle } H & & \text{angle } F \end{array}$

The third side-angle pair is side $p$ and angle $G$ . These values must be in the third ratio!

\begin{matrix} \frac{n}{\sin H} = \frac{m}{\sin F} = \frac{p}{\sin G} \end{matrix}

$\begin{array}{c} \dfrac{ n }{ \sin H } = \dfrac{ m }{ \sin F } = \dfrac{ p }{ \sin G } \end{array}$

NOTE:

We can also write the ratios upside down, but if we do this then all the ratios should match:

\begin{matrix} \frac{\sin H}{n} = \frac{\sin F}{m} = \frac{\sin G}{p} \end{matrix}

$\begin{array}{c} \dfrac{ \sin H }{ n } = \dfrac{ \sin F }{ m } = \dfrac{ \sin G }{ p } \end{array}$

The correct answer is $\dfrac{ p }{ \sin G }$ , which is choice B.

Submit your answer as:

ID is: 3504 Seed is: 2864

Finding a length with the sine rule

The diagram below shows a triangle with two angles and one side labelled. The given angles are $30°$ and $52°$ . The side across from the $30°$ angle is $\text{5}$ . Determine the length of the side across from the $52°$ angle. The figure is drawn to scale.

TIP: This diagram is drawn to scale so you can compare the answer you get to the diagram. The side you need is longer than the given side

5

$\text{5}$ . Check your answer before you submit it to make sure it agrees with the figure.

INSTRUCTION: Round your answer to two decimal places if necessary.

Answer: The length of the side is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula. Then solve for the missing side.

STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 3 / 4 points left]

The sine rule works for any triangle. It relates four pieces of information in the triangle: two angles and two sides. These angles and sides must be pairs of opposites. We can use the sine rule in either of these ways:

\begin{aligned} _{sides in denominators:}^{Angles in numerators and} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ _{angles in denominators:}^{Sides in numerators and} \frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} ^{\text{Angles in numerators and}}_{\;\text{sides in denominators:}} \quad\frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\[0.4cm] ^{\text{Sides in numerators and}}_{\;\text{angles in denominators:}} \quad\frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}$

We should pick the version that makes it easier to solve the question. To find a length, the second version above is the better choice. This is because it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So we will use this arrangement of the formula:

\begin{matrix} \frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}} \end{matrix}

$\begin{array}{c}\frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}}\end{array}$

STEP: Substitute in the known values
[−1 point ⇒ 2 / 4 points left]

When using the sine rule, we must make sure we substitute the values into the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Each of the arrows above connects numbers that belong together in one fraction. Let's use $x$ to represent the side we are looking for (the one with the ' $?$ ').

\begin{matrix} \frac{x}{\sin (52 °)} = \frac{5}{\sin (30 °)} \end{matrix}

$\begin{array}{c}\frac{x}{\sin \left(\text{52}°\right)} = \frac{ \text{5} }{\sin \left(\text{30}° \right)}\end{array}$

STEP: Solve for the missing side
[−2 points ⇒ 0 / 4 points left]

Now we can solve for $x$ . First we can evaluate everything on the right side.

\begin{aligned} \frac{x}{\sin (52 °)} & = \frac{5}{\sin (30 °)} \\ \frac{x}{\sin (52 °)} & = \frac{5}{(0.5)} \\ \frac{x}{\sin (52 °)} & = 10 \end{aligned}

$\begin{aligned} \frac{x}{\sin \left(\text{52}°\right)} & = \frac{ \text{5} }{\sin \left(\text{30}° \right)} \\ \frac{x}{\sin \left(\text{52}°\right)} & = \frac{ \text{5} }{ \left( \text{0.5} \right)} \\ \frac{x}{\sin \left(\text{52}°\right)} & = \text{10} \end{aligned}$

Next, we multiply both sides of the equation by $\sin \left(\text{52}° \right)$ to cancel the denominator on the left.

\begin{aligned} x & = (10) \cdot \sin (52 °) \\ x & = 7.88010... \\ x & \approx 7.88 \end{aligned}

$\begin{aligned} x & = \left( \text{10} \right) \cdot \sin \left(\text{52}° \right) \\ x & = \text{7.88010...} \\ x & \approx \text{7.88} \end{aligned}$

The final step rounds the answer to two decimal places, as required in the instructions. Based on the diagram, we expected the answer to be longer than than $\text{5}$ . And it is! Remember to compare your answer to the diagram if the diagram is to scale.

The length of the missing side is $\text{7.88}$ .

Submit your answer as:

ID is: 3504 Seed is: 191

Finding a length with the sine rule

The diagram below shows a triangle with two angles and one side labelled. The given angles are $104°$ and $29°$ . The side across from the $104°$ angle is $\text{8}$ . Determine the length of the side across from the $29°$ angle. The figure is drawn to scale.

TIP: This diagram is drawn to scale so you can compare the answer you get to the diagram. The side you need is shorter than the given side

8

$\text{8}$ . Check your answer before you submit it to make sure it agrees with the figure.

INSTRUCTION: Round your answer to two decimal places if necessary.

Answer: The length of the side is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula. Then solve for the missing side.

STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 3 / 4 points left]

The sine rule works for any triangle. It relates four pieces of information in the triangle: two angles and two sides. These angles and sides must be pairs of opposites. We can use the sine rule in either of these ways:

\begin{aligned} _{sides in denominators:}^{Angles in numerators and} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ _{angles in denominators:}^{Sides in numerators and} \frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} ^{\text{Angles in numerators and}}_{\;\text{sides in denominators:}} \quad\frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\[0.4cm] ^{\text{Sides in numerators and}}_{\;\text{angles in denominators:}} \quad\frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}$

We should pick the version that makes it easier to solve the question. To find a length, the second version above is the better choice. This is because it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So we will use this arrangement of the formula:

\begin{matrix} \frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}} \end{matrix}

$\begin{array}{c}\frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}}\end{array}$

STEP: Substitute in the known values
[−1 point ⇒ 2 / 4 points left]

When using the sine rule, we must make sure we substitute the values into the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Each of the arrows above connects numbers that belong together in one fraction. Let's use $x$ to represent the side we are looking for (the one with the ' $?$ ').

\begin{matrix} \frac{x}{\sin (29 °)} = \frac{8}{\sin (104 °)} \end{matrix}

$\begin{array}{c}\frac{x}{\sin \left(\text{29}°\right)} = \frac{ \text{8} }{\sin \left(\text{104}° \right)}\end{array}$

STEP: Solve for the missing side
[−2 points ⇒ 0 / 4 points left]

Now we can solve for $x$ . First we can evaluate everything on the right side.

\begin{aligned} \frac{x}{\sin (29 °)} & = \frac{8}{\sin (104 °)} \\ \frac{x}{\sin (29 °)} & = \frac{8}{(0.97029...)} \\ \frac{x}{\sin (29 °)} & = 8.24490... \end{aligned}

$\begin{aligned} \frac{x}{\sin \left(\text{29}°\right)} & = \frac{ \text{8} }{\sin \left(\text{104}° \right)} \\ \frac{x}{\sin \left(\text{29}°\right)} & = \frac{ \text{8} }{ \left( \text{0.97029...} \right)} \\ \frac{x}{\sin \left(\text{29}°\right)} & = \text{8.24490...} \end{aligned}$

Next, we multiply both sides of the equation by $\sin \left(\text{29}° \right)$ to cancel the denominator on the left.

\begin{aligned} x & = (8.24490...) \cdot \sin (29 °) \\ x & = 3.99721... \\ x & \approx 4 \end{aligned}

$\begin{aligned} x & = \left( \text{8.24490...} \right) \cdot \sin \left(\text{29}° \right) \\ x & = \text{3.99721...} \\ x & \approx \text{4} \end{aligned}$

The final step rounds the answer to two decimal places, as required in the instructions. Based on the diagram, we expected the answer to be shorter than than $\text{8}$ . And it is! Remember to compare your answer to the diagram if the diagram is to scale.

The length of the missing side is $\text{4}$ .

Submit your answer as:

ID is: 3504 Seed is: 7839

Finding a length with the sine rule

The diagram below shows a triangle with two angles and one side labelled. The given angles are $74°$ and $44°$ . The side across from the $44°$ angle is $\text{8}$ . Determine the length of the side across from the $74°$ angle. The figure is drawn to scale.

TIP: This diagram is drawn to scale so you can compare the answer you get to the diagram. The side you need is longer than the given side

8

$\text{8}$ . Check your answer before you submit it to make sure it agrees with the figure.

INSTRUCTION: Round your answer to two decimal places if necessary.

Answer: The length of the side is .

numeric

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 4 / 4 points left]

You should start by writing down the sine rule formula. Then you need to find pairs of angles and sides which are across from each other, and you can substitute them into the formula. Then solve for the missing side.

STEP: Choose which arrangement of the sine rule to use
[−1 point ⇒ 3 / 4 points left]

The sine rule works for any triangle. It relates four pieces of information in the triangle: two angles and two sides. These angles and sides must be pairs of opposites. We can use the sine rule in either of these ways:

\begin{aligned} _{sides in denominators:}^{Angles in numerators and} \frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\ _{angles in denominators:}^{Sides in numerators and} \frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}

$\begin{aligned} ^{\text{Angles in numerators and}}_{\;\text{sides in denominators:}} \quad\frac{\sin \hat{A}}{a} & = \frac{\sin \hat{B}}{b} \\[0.4cm] ^{\text{Sides in numerators and}}_{\;\text{angles in denominators:}} \quad\frac{a}{\sin \hat{A}} & = \frac{b}{\sin \hat{B}} \end{aligned}$

We should pick the version that makes it easier to solve the question. To find a length, the second version above is the better choice. This is because it is easier to solve equations when the unknown variable is in the numerator of a fraction (instead of the denominator). So we will use this arrangement of the formula:

\begin{matrix} \frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}} \end{matrix}

$\begin{array}{c}\frac{a}{\sin \hat{A}} = \frac{b}{\sin \hat{B}}\end{array}$

STEP: Substitute in the known values
[−1 point ⇒ 2 / 4 points left]

When using the sine rule, we must make sure we substitute the values into the correct places. Each fraction must have an angle and the side of the triangle opposite from each other. It can be helpful to circle and connect these opposite pairs, like this:

Each of the arrows above connects numbers that belong together in one fraction. Let's use $x$ to represent the side we are looking for (the one with the ' $?$ ').

\begin{matrix} \frac{x}{\sin (74 °)} = \frac{8}{\sin (44 °)} \end{matrix}

$\begin{array}{c}\frac{x}{\sin \left(\text{74}°\right)} = \frac{ \text{8} }{\sin \left(\text{44}° \right)}\end{array}$

STEP: Solve for the missing side
[−2 points ⇒ 0 / 4 points left]

Now we can solve for $x$ . First we can evaluate everything on the right side.

\begin{aligned} \frac{x}{\sin (74 °)} & = \frac{8}{\sin (44 °)} \\ \frac{x}{\sin (74 °)} & = \frac{8}{(0.69465...)} \\ \frac{x}{\sin (74 °)} & = 11.51645... \end{aligned}

$\begin{aligned} \frac{x}{\sin \left(\text{74}°\right)} & = \frac{ \text{8} }{\sin \left(\text{44}° \right)} \\ \frac{x}{\sin \left(\text{74}°\right)} & = \frac{ \text{8} }{ \left( \text{0.69465...} \right)} \\ \frac{x}{\sin \left(\text{74}°\right)} & = \text{11.51645...} \end{aligned}$

Next, we multiply both sides of the equation by $\sin \left(\text{74}° \right)$ to cancel the denominator on the left.

\begin{aligned} x & = (11.51645...) \cdot \sin (74 °) \\ x & = 11.07032... \\ x & \approx 11.07 \end{aligned}

$\begin{aligned} x & = \left( \text{11.51645...} \right) \cdot \sin \left(\text{74}° \right) \\ x & = \text{11.07032...} \\ x & \approx \text{11.07} \end{aligned}$

The final step rounds the answer to two decimal places, as required in the instructions. Based on the diagram, we expected the answer to be longer than than $\text{8}$ . And it is! Remember to compare your answer to the diagram if the diagram is to scale.

The length of the missing side is $\text{11.07}$ .

Submit your answer as:

ID is: 3558 Seed is: 6275

Algebra meets the sine rule

In triangle $EFD$ below, two angles are given: $\hat{E} = 61°$ and $\hat{D} = 70°$ . Side $\overline{DE}$ and $\overline{FD}$ have lengths $3 x + 9$ $\text{cm}$ and $7 x - 1$ $\text{cm}$ respectively, where $x \in \mathbb{R}$ . The third side, $\overline{EF}$ , is $\text{11}\ \text{cm}$ long. The figure may or may not be drawn to scale.

Determine the value of $x$ .

INSTRUCTION: Round your answer to two decimal places.

Answer: $x =$
one-of
type(numeric.abserror(0.004))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing down an equation using the sine rule. You should make sure that your equation includes side $\overline{FD}$ , which is equal to $7 x - 1$ .

STEP: Write a sine rule equation including $x$
[−1 point ⇒ 2 / 3 points left]

We need to determine the value of $x$ . We can do this by using the sine rule:

$\begin{array}{c} \dfrac{7 x - 1}{\sin \left(61°\right)} = \dfrac{\text{11}}{\sin \left(70°\right)} \end{array}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation. Start by multiplying both sides of the equation by $\sin \left(61°\right)$ . Then work out all the calculations on the right side.

$\begin{aligned} \frac{7 x - 1}{\sin \left(61°\right)} & = \frac{\text{11}}{\sin \left(70°\right)} \\ 7 x - 1 & = \frac{\text{11}}{\sin 70°} \cdot \left(\sin 61°\right) \\ 7 x - 1 & = \frac{\text{11}}{\text{0.9396...}} \cdot \left(\text{0.8746...}\right) \\ 7 x - 1 & = \text{10.2382...} \\ x & = \frac{ \text{10.2382...} +1 }{ 7 } \\ x & = \text{1.6054...} \\ & \approx \text{1.61} \end{aligned}$

Rounding to two decimal places, the value of $x$ is $\text{1.61}$ .

Submit your answer as:
Hence determine the length of $\overline{DE}$ .
INSTRUCTION:
- Use the rounded answer from Question 1.
- Round your answer to two decimal places if necessary.
Answer: $\overline{DE} =$ $\text{cm}$
one-of
type(numeric.noerror)
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Substitute the answer to Question 1 into the expression for $\overline{DE}$ , which is $3 x + 9$ .

STEP: Substitute and evaluate
[−1 point ⇒ 0 / 1 points left]

We need to find the length of side $\overline{DE}$ . It has a length of $3 x + 9$ . So we can substitute in the answer from Question 1 and work out the length.

$\begin{aligned} \overline{DE} & = 3 x + 9 \\ & = 3 \left(\text{1.61}\right) + 9 \\ & = \text{13.83}\ \text{cm} \end{aligned}$

This value represents a distance (one side of a triangle) so it cannot be negative. Since the answer is positive, everything looks good. (Remember to check your answers if the answer represents a real thing: the answer must be realistic.)

The length of $\overline{DE}$ is $\text{13.83}\ \text{cm}$ .

Submit your answer as:

ID is: 3558 Seed is: 5463

Algebra meets the sine rule

In triangle $ACB$ below, two angles are given: $\hat{A} = 86°$ and $\hat{B} = 67°$ . Side $\overline{BA}$ and $\overline{CB}$ have lengths $3 y - 4$ $\text{cm}$ and $2 y - 4$ $\text{cm}$ respectively, where $y \in \mathbb{R}$ . The third side, $\overline{AC}$ , is $\text{11}\ \text{cm}$ long. The figure may or may not be drawn to scale.

Determine the value of $y$ .

INSTRUCTION: Round your answer to two decimal places.

Answer: $y =$
one-of
type(numeric.abserror(0.004))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing down an equation using the sine rule. You should make sure that your equation includes side $\overline{CB}$ , which is equal to $2 y - 4$ .

STEP: Write a sine rule equation including $y$
[−1 point ⇒ 2 / 3 points left]

We need to determine the value of $y$ . We can do this by using the sine rule:

$\begin{array}{c} \dfrac{2 y - 4}{\sin \left(86°\right)} = \dfrac{\text{11}}{\sin \left(67°\right)} \end{array}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation. Start by multiplying both sides of the equation by $\sin \left(86°\right)$ . Then work out all the calculations on the right side.

$\begin{aligned} \frac{2 y - 4}{\sin \left(86°\right)} & = \frac{\text{11}}{\sin \left(67°\right)} \\ 2 y - 4 & = \frac{\text{11}}{\sin 67°} \cdot \left(\sin 86°\right) \\ 2 y - 4 & = \frac{\text{11}}{\text{0.9205...}} \cdot \left(\text{0.9975...}\right) \\ 2 y - 4 & = \text{11.9208...} \\ y & = \frac{ \text{11.9208...} +4 }{ 2 } \\ y & = \text{7.9604...} \\ & \approx \text{7.96} \end{aligned}$

Rounding to two decimal places, the value of $y$ is $\text{7.96}$ .

Submit your answer as:
Hence determine the length of $\overline{BA}$ .
INSTRUCTION:
- Use the rounded answer from Question 1.
- Round your answer to two decimal places if necessary.
Answer: $\overline{BA} =$ $\text{cm}$
one-of
type(numeric.noerror)
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Substitute the answer to Question 1 into the expression for $\overline{BA}$ , which is $3 y - 4$ .

STEP: Substitute and evaluate
[−1 point ⇒ 0 / 1 points left]

We need to find the length of side $\overline{BA}$ . It has a length of $3 y - 4$ . So we can substitute in the answer from Question 1 and work out the length.

$\begin{aligned} \overline{BA} & = 3 y - 4 \\ & = 3 \left(\text{7.96}\right) - 4 \\ & = \text{19.88}\ \text{cm} \end{aligned}$

This value represents a distance (one side of a triangle) so it cannot be negative. Since the answer is positive, everything looks good. (Remember to check your answers if the answer represents a real thing: the answer must be realistic.)

The length of $\overline{BA}$ is $\text{19.88}\ \text{cm}$ .

Submit your answer as:

ID is: 3558 Seed is: 3544

Algebra meets the sine rule

In triangle $FED$ below, two angles are given: $\hat{F} = 72°$ and $\hat{D} = 62°$ . Side $\overline{DF}$ and $\overline{ED}$ have lengths $4 y - 7$ $\text{cm}$ and $3 y + 3$ $\text{cm}$ respectively, where $y \in \mathbb{R}$ . The third side, $\overline{FE}$ , is $\text{16}\ \text{cm}$ long. The figure may or may not be drawn to scale.

What is the value of $y$ ?

INSTRUCTION: Round your answer to two decimal places.

Answer: $y =$
one-of
type(numeric.abserror(0.004))
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing down an equation using the sine rule. You should make sure that your equation includes side $\overline{ED}$ , which is equal to $3 y + 3$ .

STEP: Write a sine rule equation including $y$
[−1 point ⇒ 2 / 3 points left]

We need to determine the value of $y$ . We can do this by using the sine rule:

$\begin{array}{c} \dfrac{3 y + 3}{\sin \left(72°\right)} = \dfrac{\text{16}}{\sin \left(62°\right)} \end{array}$

STEP: Solve the equation
[−2 points ⇒ 0 / 3 points left]

Now we can solve the equation. Start by multiplying both sides of the equation by $\sin \left(72°\right)$ . Then work out all the calculations on the right side.

$\begin{aligned} \frac{3 y + 3}{\sin \left(72°\right)} & = \frac{\text{16}}{\sin \left(62°\right)} \\ 3 y + 3 & = \frac{\text{16}}{\sin 62°} \cdot \left(\sin 72°\right) \\ 3 y + 3 & = \frac{\text{16}}{\text{0.8829...}} \cdot \left(\text{0.9510...}\right) \\ 3 y + 3 & = \text{17.2342...} \\ y & = \frac{ \text{17.2342...} -3 }{ 3 } \\ y & = \text{4.7447...} \\ & \approx \text{4.74} \end{aligned}$

Rounding to two decimal places, the value of $y$ is $\text{4.74}$ .

Submit your answer as:
Hence determine the length of $\overline{ED}$ .
INSTRUCTION:
- Use the rounded answer from Question 1.
- Round your answer to two decimal places if necessary.
Answer: $\overline{ED} =$ $\text{cm}$
one-of
type(numeric.noerror)
2 attempts remaining

HINT: <no title>
[−0 points ⇒ 1 / 1 points left]

Substitute the answer to Question 1 into the expression for $\overline{ED}$ , which is $3 y + 3$ .

STEP: Substitute and evaluate
[−1 point ⇒ 0 / 1 points left]

We need to find the length of side $\overline{ED}$ . It has a length of $3 y + 3$ . So we can substitute in the answer from Question 1 and work out the length.

$\begin{aligned} \overline{ED} & = 3 y + 3 \\ & = 3 \left(\text{4.74}\right) + 3 \\ & = \text{17.22}\ \text{cm} \end{aligned}$

This value represents a distance (one side of a triangle) so it cannot be negative. Since the answer is positive, everything looks good. (Remember to check your answers if the answer represents a real thing: the answer must be realistic.)

The length of $\overline{ED}$ is $\text{17.22}\ \text{cm}$ .

Submit your answer as:

ID is: 3570 Seed is: 2079

Sine rule problems

In the figure below, $W\hat{Y}X = m$ , $Y\hat{X}W = n$ , and $YW = d$ . $XY$ and $WZ$ are perpendicular to $ZX$ .

Determine the distance $XY$ in terms of $m$ , $n$ , and $d$ .

Answer:

X Y =

$XY =$

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation about triangle $YWX$ . You need this equation to include $m$ , $n$ , and $d$ . So focus on the sides and angles which will bring those variables into the equation.

STEP: Add any information you can to the diagram
[−1 point ⇒ 2 / 3 points left]

We need to find an equation for $XY$ in terms of the three variables $m$ , $n$ , and $d$ . So we need to write an equation about the diagram which includes all four of those things.

How to do that is not obvious because the diagram is a bit complex. Which triangle should we use? Can we take advantage of the right-angles? It is not easy to know at the beginning. So as a first step, we can examine the diagram and label any extra information we can find in the figure.

In this case, we can know four new pieces of information:

$XY$ and $WZ$ must be parallel to each other because they are both perpendicular to $ZX$ .
$Z\hat{X}W = 90° - n$ because it is complementary to angle $n$ .
$Z\hat{W}X = n$ because it is an alternate interior angle with $n$ ( $XY$ and $WZ$ are parallel).
$Y\hat{W}X$ must be $180° - (m + n)$ because the angles in $\Delta YWX$ add up to $180°$ .

At this point we do not know if any of that information will be useful. We will find out!

STEP: Write an equation and try to include

m

$m$ ,

n

$n$ , and

d

$d$
[−1 point ⇒ 1 / 3 points left]

Now let's write an equation about the diagram. We need an equation which includes the variables $m$ , $n$ , and $d$ (because those must be in the answer). So we need to focus on the sides and angles which will bring those variables into the equation. The equation must also include side $XY$ .

We can try the sine rule. We can get everything we need into the equation if we focus on the sides and angles in $\Delta YWX$ .

\begin{matrix} \frac{X Y}{\sin (180 ° - (m + n))} = \frac{d}{\sin n} \end{matrix}

$\begin{array}{c} \dfrac{XY}{\sin \left(180° - (m + n)\right)} = \dfrac{d}{\sin n} \end{array}$

Now we are in business: the answer must be in terms of $m$ , $n$ . And those are the variables in the equation (in addition to $XY$ ).

STEP: Simplify the equation to get the expected form
[−1 point ⇒ 0 / 3 points left]

We need to solve the equation for $XY$ . To do this, multiply both sides of the equation by $\sin \left(180° - (m + n)\right)$ .

\begin{aligned} X Y & = \frac{d \cdot \sin (180 ° - (m + n))}{\sin n} \end{aligned}

$\begin{aligned} XY & = \frac{d \cdot \sin \left(180° - (m + n)\right)}{\sin n} \end{aligned}$

Now we can use a trigonometric identity to simplify this equation. We know that $\sin (180° - A)$ is equal to $\sin A$ . So the quantity $\sin (180° - (m + n))$ is equal to $\sin (m + n)$ !

\begin{aligned} X Y & = \frac{d \cdot \sin (180 ° - (m + n))}{\sin n} \\ X Y & = \frac{d \cdot \sin (m + n)}{\sin n} \end{aligned}

$\begin{aligned} XY & = \frac{d \cdot \sin \left(180° - (m + n)\right)}{\sin n} \\[0.2cm] XY & = \frac{d \cdot \sin (m + n)}{\sin n} \end{aligned}$

That gives us a tidy and accurate expression for $XY$ in terms of the required values.

The length of $XY$ , as a function of $m$ , $n$ , and $d$ is $\dfrac{d \cdot \sin (m + n)}{\sin n}$ .

Submit your answer as:

ID is: 3570 Seed is: 7555

Sine rule problems

In the figure below, $Z\hat{X}W = f$ , $X\hat{W}Z = g$ , and $XZ = d$ . $WX$ and $ZY$ are perpendicular to $YW$ .

Determine the distance $WX$ in terms of $f$ , $g$ , and $d$ .

Answer:

W X =

$WX =$

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation about triangle $XZW$ . You need this equation to include $f$ , $g$ , and $d$ . So focus on the sides and angles which will bring those variables into the equation.

STEP: Add any information you can to the diagram
[−1 point ⇒ 2 / 3 points left]

We need to find an equation for $WX$ in terms of the three variables $f$ , $g$ , and $d$ . So we need to write an equation about the diagram which includes all four of those things.

How to do that is not obvious because the diagram is a bit complex. Which triangle should we use? Can we take advantage of the right-angles? It is not easy to know at the beginning. So as a first step, we can examine the diagram and label any extra information we can find in the figure.

In this case, we can know four new pieces of information:

$WX$ and $ZY$ must be parallel to each other because they are both perpendicular to $YW$ .
$Y\hat{W}Z = 90° - g$ because it is complementary to angle $g$ .
$Y\hat{Z}W = g$ because it is an alternate interior angle with $g$ ( $WX$ and $ZY$ are parallel).
$X\hat{Z}W$ must be $180° - (f + g)$ because the angles in $\Delta XZW$ add up to $180°$ .

At this point we do not know if any of that information will be useful. We will find out!

STEP: Write an equation and try to include

f

$f$ ,

g

$g$ , and

d

$d$
[−1 point ⇒ 1 / 3 points left]

Now let's write an equation about the diagram. We need an equation which includes the variables $f$ , $g$ , and $d$ (because those must be in the answer). So we need to focus on the sides and angles which will bring those variables into the equation. The equation must also include side $WX$ .

We can try the sine rule. We can get everything we need into the equation if we focus on the sides and angles in $\Delta XZW$ .

\begin{matrix} \frac{W X}{\sin (180 ° - (f + g))} = \frac{d}{\sin g} \end{matrix}

$\begin{array}{c} \dfrac{WX}{\sin \left(180° - (f + g)\right)} = \dfrac{d}{\sin g} \end{array}$

Now we are in business: the answer must be in terms of $f$ , $g$ . And those are the variables in the equation (in addition to $WX$ ).

STEP: Simplify the equation to get the expected form
[−1 point ⇒ 0 / 3 points left]

We need to solve the equation for $WX$ . To do this, multiply both sides of the equation by $\sin \left(180° - (f + g)\right)$ .

\begin{aligned} W X & = \frac{d \cdot \sin (180 ° - (f + g))}{\sin g} \end{aligned}

$\begin{aligned} WX & = \frac{d \cdot \sin \left(180° - (f + g)\right)}{\sin g} \end{aligned}$

Now we can use a trigonometric identity to simplify this equation. We know that $\sin (180° - A)$ is equal to $\sin A$ . So the quantity $\sin (180° - (f + g))$ is equal to $\sin (f + g)$ !

\begin{aligned} W X & = \frac{d \cdot \sin (180 ° - (f + g))}{\sin g} \\ W X & = \frac{d \cdot \sin (f + g)}{\sin g} \end{aligned}

$\begin{aligned} WX & = \frac{d \cdot \sin \left(180° - (f + g)\right)}{\sin g} \\[0.2cm] WX & = \frac{d \cdot \sin (f + g)}{\sin g} \end{aligned}$

That gives us a tidy and accurate expression for $WX$ in terms of the required values.

The length of $WX$ , as a function of $f$ , $g$ , and $d$ is $\dfrac{d \cdot \sin (f + g)}{\sin g}$ .

Submit your answer as:

ID is: 3570 Seed is: 4720

Sine rule problems

In the figure below, $R\hat{P}Q = f$ , $P\hat{Q}R = g$ , and $QP = h$ . $QP$ and $RS$ are perpendicular to $SQ$ .

Determine the distance $RQ$ in terms of $f$ , $g$ , and $h$ .

Answer:

R Q =

$RQ =$

expression

2 attempts remaining

HINT: <no title>
[−0 points ⇒ 3 / 3 points left]

Start by writing an equation about triangle $PRQ$ . You need this equation to include $f$ , $g$ , and $h$ . So focus on the sides and angles which will bring those variables into the equation.

STEP: Add any information you can to the diagram
[−1 point ⇒ 2 / 3 points left]

We need to find an equation for $RQ$ in terms of the three variables $f$ , $g$ , and $h$ . So we need to write an equation about the diagram which includes all four of those things.

How to do that is not obvious because the diagram is a bit complex. Which triangle should we use? Can we take advantage of the right-angles? It is not easy to know at the beginning. So as a first step, we can examine the diagram and label any extra information we can find in the figure.

In this case, we can know four new pieces of information:

$QP$ and $RS$ must be parallel to each other because they are both perpendicular to $SQ$ .
$S\hat{Q}R = 90° - g$ because it is complementary to angle $g$ .
$S\hat{R}Q = g$ because it is an alternate interior angle with $g$ ( $QP$ and $RS$ are parallel).
$P\hat{R}Q$ must be $180° - (f + g)$ because the angles in $\Delta PRQ$ add up to $180°$ .

At this point we do not know if any of that information will be useful. We will find out!

STEP: Write an equation and try to include

f

$f$ ,

g

$g$ , and

h

$h$
[−1 point ⇒ 1 / 3 points left]

Now let's write an equation about the diagram. We need an equation which includes the variables $f$ , $g$ , and $h$ (because those must be in the answer). So we need to focus on the sides and angles which will bring those variables into the equation. The equation must also include side $RQ$ .

We can try the sine rule. We can get everything we need into the equation if we focus on the sides and angles in $\Delta PRQ$ .

\begin{matrix} \frac{R Q}{\sin f} = \frac{h}{\sin (180 ° - (f + g))} \end{matrix}

$\begin{array}{c} \dfrac{RQ}{\sin f} = \dfrac{h}{\sin \left(180° - (f + g)\right)} \end{array}$

Now we are in business: the answer must be in terms of $f$ , $g$ . And those are the variables in the equation (in addition to $RQ$ ).

STEP: Simplify the equation to get the expected form
[−1 point ⇒ 0 / 3 points left]

We need to solve the equation for $RQ$ . To do this, multiply both sides of the equation by $\sin f$ .

\begin{aligned} R Q & = \frac{h \cdot \sin f}{\sin (180 ° - (f + g))} \end{aligned}

$\begin{aligned} RQ & = \frac{h \cdot \sin f}{\sin \left(180° - (f + g)\right)} \end{aligned}$

Now we can use a trigonometric identity to simplify this equation. We know that $\sin (180° - A)$ is equal to $\sin A$ . So the quantity $\sin (180° - (f + g))$ is equal to $\sin (f + g)$ !

\begin{aligned} R Q & = \frac{h \cdot \sin f}{\sin (180 ° - (f + g))} \\ R Q & = \frac{h \cdot \sin f}{\sin (f + g)} \end{aligned}

$\begin{aligned} RQ & = \frac{h \cdot \sin f}{\sin \left(180° - (f + g)\right)} \\[0.2cm] RQ & = \frac{h \cdot \sin f}{\sin (f + g)} \end{aligned}$

That gives us a tidy and accurate expression for $RQ$ in terms of the required values.

The length of $RQ$ , as a function of $f$ , $g$ , and $h$ is $\dfrac{h \cdot \sin f}{\sin (f + g)}$ .

Submit your answer as:

Proof Steps
1.	$= \dfrac{x}{p}$
2.	$\therefore x = -p\cos{\hat{R}}$
	In $\Delta QPS$
3.	$r^{2} \;=\; QS^{2} \;+\;$ $\;$ $(\text{Pythagoras})$
4.	$\therefore QS^{2} = \;$ $r^{2} - q^{2} - 2qx - x^{2}$
	In $\Delta QSR$
5.	$p^{2} = QS^{2} + x^{2} \;$ $(\text{Pythagoras})$
6.	$\therefore QS^{2} =$
7.	$r^{2} - q^{2} - 2qx - x^{2}$ $\;=\;$ $p^{2} - x^{2}$
8.	$r^{2} = q^{2} + p^{2} +$
9.	$\therefore r^{2} =$ $\;p^{2} + q^{2}$ $\; -\; 2pq\cos{\hat{R}}$

Proof Steps
1.	$\cos{(\text{180}\text{°} - \hat{R})} = \dfrac{x}{p}$
2.	$\therefore x = -p\cos{\hat{R}}$
	In $\Delta QPS$
3.	$r^{2} = QS^{2} + (q + x)^{2} \;$ $(\text{Pythagoras})$
4.	$\therefore QS^{2} =\;$ $r^{2} - q^{2} - 2qx - x^{2}$
	In $\Delta QSR$
5.	$p^{2} = QS^{2} + x^{2} \;$ $(\text{Pythagoras})$
6.	$\therefore QS^{2} = p^{2} - x^{2}$
7.	$r^{2} - q^{2} - 2qx - x^{2}$ $\;=\;$ $p^{2} - x^{2}$
8.	$r^{2} = q^{2} + p^{2} + 2qx$
9.	$\therefore r^{2} =$ $\;p^{2} + q^{2}$ $\; -\; 2pq\cos{\hat{R}}$

Proof Steps
1.	$A(b; 0) \;$ $B($ $; a\sin{\hat{C}}) \;$ $C(0; 0)$
2.	$AB^{2} = AD^{2} + BD^{2} \;$ $(\text{Pythagoras})$
3.	$c^{2} =$ $\; (b - a\cos{\hat{C}})^{2} \;+\;$
4.	$c^{2} =$ $\; b^{2} - 2ab\cos{\hat{C}} + a^{2}\cos^{2}{\hat{C}} \;+\;$
5.	$c^{2} =$ $\; b^{2} - 2ab\cos{\hat{C}}$ $\;+\; a^{2}(\cos^{2}{\hat{C}} + \sin^{2}{\hat{C}})$
6.	$c^{2} =$ $\;b^{2}$ $\;-\; 2ab\cos{\hat{C}}$ $\;+\; a^{2}($ $)$
7.	$\therefore c^{2} =$ $\;a^{2} + b^{2}$ $\; -\; 2ab\cos{\hat{C}}$

Proof Steps
1.	$A(b; 0)$ $\; B(a\cos{\hat{C}}; a\sin{\hat{C}})$ $\; C(0; 0)$
2.	$AB^{2} = AD^{2} + BD^{2} \;$ $(\text{Pythagoras})$
3.	$c^{2} =$ $\; (b - a\cos{\hat{C}})^{2}$ $\; + \; (0 - a\sin{\hat{C}})^{2}$
4.	$c^{2} =$ $\; b^{2}$ $\;-\; 2ab\cos{\hat{C}}$ $\;+\; a^{2}\cos^{2}{\hat{C}}$ $\;+\; a^{2}\sin^{2}{\hat{C}}$
5.	$c^{2} = \;$ $b^{2}$ $\;-\; 2ab\cos{\hat{C}}$ $\;+\; a^{2}(\cos^{2}{\hat{C}}$ $\;+\; \sin^{2}{\hat{C}})$
6.	$c^{2} =$ $\;b^{2}$ $\;-\; 2ab\cos{\hat{C}}$ $\;+\; a^{2}(1)$
7.	$\therefore c^{2} =$ $\;a^{2} + b^{2}$ $\; -\; 2ab\cos{\hat{C}}$

Proof Steps
1.	$P(q; 0) \;$ $Q($ $; p\sin{\hat{R}}) \;$ $R(0; 0)$
2.	$PQ^{2} = PS^{2} + QS^{2} \;$ $(\text{Pythagoras})$
3.	$r^{2} =$ $\; (q - p\cos{\hat{R}})^{2} \;+\;$
4.	$r^{2} =$ $\; q^{2} - 2pq\cos{\hat{R}} + p^{2}\cos^{2}{\hat{R}} \;+\;$
5.	$r^{2} =$ $\; q^{2} - 2pq\cos{\hat{R}}$ $\;+\; p^{2}(\cos^{2}{\hat{R}} + \sin^{2}{\hat{R}})$
6.	$r^{2} =$ $\;q^{2}$ $\;-\; 2pq\cos{\hat{R}}$ $\;+\; p^{2}($ $)$
7.	$\therefore r^{2} =$ $\;p^{2} + q^{2}$ $\; -\; 2pq\cos{\hat{R}}$

Proof Steps
1.	$P(q; 0)$ $\; Q(p\cos{\hat{R}}; p\sin{\hat{R}})$ $\; R(0; 0)$
2.	$PQ^{2} = PS^{2} + QS^{2} \;$ $(\text{Pythagoras})$
3.	$r^{2} =$ $\; (q - p\cos{\hat{R}})^{2}$ $\; + \; (0 - p\sin{\hat{R}})^{2}$
4.	$r^{2} =$ $\; q^{2}$ $\;-\; 2pq\cos{\hat{R}}$ $\;+\; p^{2}\cos^{2}{\hat{R}}$ $\;+\; p^{2}\sin^{2}{\hat{R}}$
5.	$r^{2} = \;$ $q^{2}$ $\;-\; 2pq\cos{\hat{R}}$ $\;+\; p^{2}(\cos^{2}{\hat{R}}$ $\;+\; \sin^{2}{\hat{R}})$
6.	$r^{2} =$ $\;q^{2}$ $\;-\; 2pq\cos{\hat{R}}$ $\;+\; p^{2}(1)$
7.	$\therefore r^{2} =$ $\;p^{2} + q^{2}$ $\; -\; 2pq\cos{\hat{R}}$

Choices	If you use a $90°$ angle is the cosine rule...
A	the sum of the interior angles of a triangle is $180°$
B	it becomes the theorem of Pythagoras
C	it changes into the sine rule
D	it changes into the area rule

Choices	If you use a $90°$ angle is the cosine rule...
A	it becomes the equilateral triangle calculation
B	it becomes the theorem of Pythagoras
C	it changes into the sine rule
D	the sum of the interior angles of a triangle is $180°$

Choices	If you use a $90°$ angle is the cosine rule...
A	it becomes the isosceles triangle calculation
B	it changes into the sine rule
C	it becomes the theorem of Pythagoras
D	it changes into the area rule

I	II

III	IV

I	II

III	IV

I	II

III	IV

Trigonometry

1. Sine rule

The sine rule: finding mistakes

The sine rule: finding mistakes

The sine rule: finding mistakes

Using the sine rule

Using the sine rule

Using the sine rule

Working algebraically: finding a sine ratio

Working algebraically: finding a sine ratio

Working algebraically: finding a sine ratio

Finding an angle with the sine rule

Finding an angle with the sine rule

Finding an angle with the sine rule

The flexibility of the sine rule

The flexibility of the sine rule

The flexibility of the sine rule

The sine rule: identifying the correct sides and angles

The sine rule: identifying the correct sides and angles

The sine rule: identifying the correct sides and angles

Find the missing length

Find the missing length

Find the missing length

Connecting the sine rule to right triangles

Connecting the sine rule to right triangles

Connecting the sine rule to right triangles

The flexibility of the sine rule

The flexibility of the sine rule

The flexibility of the sine rule

Finding a length with the sine rule

Finding a length with the sine rule

Finding a length with the sine rule

Algebra meets the sine rule

Algebra meets the sine rule

Algebra meets the sine rule

Sine rule problems

Sine rule problems

Sine rule problems

2. Cosine rule

Perimeter problems

Perimeter problems

Perimeter problems

Application of coscos\cos rule to prove an identity

Application of coscos\cos rule to prove an identity

Application of coscos\cos rule to prove an identity

Figuring out what information to use

Figuring out what information to use

Figuring out what information to use

Finding the side opposite a known angle

Finding the side opposite a known angle

Finding the side opposite a known angle

The cosine rule: finding mistakes

The cosine rule: finding mistakes

The cosine rule: finding mistakes

Cos rule proof

Cos rule proof

Cos rule proof

Using the cosine rule: is there enough information?

Using the cosine rule: is there enough information?

Using the cosine rule: is there enough information?

Finding an angle with the cosine rule

Finding an angle with the cosine rule

Finding an angle with the cosine rule

Using the cosine rule

Using the cosine rule

Using the cosine rule

The cosine rule: identifying the correct sides and angles

The cosine rule: identifying the correct sides and angles

The cosine rule: identifying the correct sides and angles

Cosine rule connections

Cosine rule connections

Cosine rule connections

Finding an adjacent side

Finding an adjacent side

Finding an adjacent side

3. Practical applications

Predator and prey

Predator and prey

Predator and prey

Using the sinsin\sin rule & coscos\cos rule to find unknown sides & angles

Application of $\cos$ rule to prove an identity

Application of $\cos$ rule to prove an identity

Application of $\cos$ rule to prove an identity

Using the $\sin$ rule & $\cos$ rule to find unknown sides & angles

Using the $\sin$ rule & $\cos$ rule to find unknown sides & angles

Using the $\sin$ rule & $\cos$ rule to find unknown sides & angles